In $\triangle ABC$ and $\triangle PQR$ , $BC=QR, \angle A=90^{\circ},\angle C =\angle R=40^{\circ}\ and\ \angle Q=50^{\circ}.$
Above two triangles are congruent by ASA test
If true then enter $1$ else if False enter $0.$

$\triangle PQR$

$\angle P + \angle Q + \angle R = 180$

$\angle P + 40 + 50 = 180$

$\angle P = 90$

Now, in

$\triangle$ s,

$ABC$ and

$PQR,$

$BC = QR$ (Given)

$\angle A = \angle P = 90^{\circ}$

$\angle C = \angle R = 40^{\circ}$ (Given)
Thus,

$\triangle ABC \cong \triangle PQR$ (ASA postulate)

The given statement is true.

In $\bigtriangleup ABC \ and \bigtriangleup PQR, AB=PQ,AC=PR\:and\:BC =QR$ , then the two triangles are congruent by which test?

$\triangle$ s, ABC and PQR,
It is given,

$AB = PQ$

$AC = PR$

$BC = QR$
Thus,

$\triangle ABC \cong \triangle PQR$ (SSS postulate)

In $\Delta$ PQR, seg $PQ$ $\cong$ seg $QR$ and $\angle P = 70^o$ , then find the measures of angle $Q$ is:

Given, seg

$PQ$

$\cong$ seg

$QR$ .
Thus,

$\angle R = \angle P = 70^{\circ}$ (Isosceles triangle property).

We know, sum of angles of the triangle

$= 180^o$ .
Then,

$\angle P + \angle Q + \angle R = 180$

$\implies$

$70 + 70 + \angle Q = 180$

$\implies$

$\angle Q = 40^{\circ}$ .

Without drawing exact triangles, state whether the given pairs of triangle are congruent or not, if congruent use '1' else '0'.
In $\Delta ABC \,\,\, and \,\,\, \Delta DEF; AB=EF, BC=DF \,\, and \,\, \angle B = \angle F$

$\triangle$ , ABC and DEF

$AB = EF$ (Given)

$BC = DF$ (Given)

$\angle B = \angle F$ (Given)

Thus,

$\triangle ABC \cong \triangle DEF$ (By SAS rule)

In $\Delta ABC,$ if $AB = 76$ cm, $BC=69$ cm and $CA=61$ cm, then the smallest angle is $B$ .
If true then enter $1$ and if false then enter $0$ .

Given:

$AB = 76$ ,

$BC = 69$ and

$CA = 61$
Since,

$CA$ is the smallest, so the angle opposite to

$CA$ will be the smallest. Hence,

$\angle B$ is the smallest angle.

In the alongside diagram, $ABCD$ is parallelogram. Prove that : $AB = 2 BC$

Given,

$ABCD$ is a parallelogram

The bisectors of

$\angle A,$ bisect

$DC$ at

$E$

$\therefore AB||CD$ [opposite sides of a parallelogram are equal]

$AB||CD$ and

$AE$ is transversal

$\angle BAE=\angle AED$ [alternate angle]....(i)

$$\angle DAE =\angle BAE [AE is the bisector of \angle A]....(ii)

From (i) and (ii),

$\angle AED=\angle DAE$

$\therefore$ in

$\triangle DAE$

$\angle AED=\angle DAE$

$\Rightarrow AD=DE$ [If two angles of a triangle are equal, then sides opposite to them are equal.]

$2AD=2DE=DE+DE=DE+EC$

$\Rightarrow 2AD=DC$

$AD=BC$ and

$DC=AB \Rightarrow AB=2BC$

Without drawing exact triangles, state, giving reasons, whether the given pairs of triangle are congruent or not :
$\Delta PQR \,\,\, and \,\,\, \Delta SQR; PQ=SR\,\, and \,\, \angle PQR = \angle SQR$ .

$\triangle$ PQR and SQR,

$PQ = SR$ (Given)

$\angle PQR = \angle SQR$ (Given)

$QR = QR$ (Common)

Hence,

$\triangle PQR \cong \triangle SQR$ (SAS postulate)

Without drawing exact triangles, state, giving reasons, whether the given pairs of triangle are congruent or not :
In $\Delta ABC \ and \ \Delta PQR; AB=QR, AC=PR \,\, and \,\, \angle B = \angle R$ .

$\triangle ABC$ and

$\triangle PQR$ ,

$AB = QR$ (Given)

$AC = PR$ (Given)

$\angle B = \angle R$ (Given)

Corresponding pairs of both the triangles are not equal. The two triangles thus are not congruent.

If the two sides and the included angle of a triangle are respectively equal to two sides and the included angle of the other triangle, the two triangles are congruent. If true enter 1 else 0.

If two sides and the included angle of a triangle are respectively equal to two sides and the included angle of the other triangle, then the two triangles are congruent by the SAS rule.

If $\triangle DEF\cong \triangle RPQ$ , then $\angle D=\angle Q$ .
If true enter $1$ , else $0$ .

Given $\triangle DEF \cong \triangle RPQ$ .

Then the corresponding parts will also be equal.

That is by CPCT rule, corresponding parts of congruent triangles are equal.

Then, $DE=RP$ , $EF=PQ$ , $DF=RQ$ , $\angle D=\angle R$ , $\angle E=\angle P$ and $\angle F=\angle Q$ .

Thus, $\angle D\ne\angle Q$ .

Therefore, the statement is false and $0$ is the answer.

If $\triangle PQR\cong \triangle CAB$ , $PQ = CA$ .
If true enter $1$ , else $0$ .

Given:

$\triangle PQR \cong \triangle CAB$

Then the corresponding parts will also be equal.

That is by CPCT rule, corresponding parts of congruent triangles are equal.

Then, $PQ = CA$ , $QR=AB$ , $PR=CB$ , $\angle P=\angle C$ , $\angle Q=\angle A$ and $\angle R=\angle B$ .

Thus,

$PQ = CA$ .
Therefore, the statement is true and

$1$ is the answer

If $\triangle ABC\cong \triangle PRQ$ then $AB = PQ$ .
If true enter $1$ , else $0$ .

Given

$\triangle ABC \cong \triangle PRQ$ .

Then the corresponding parts will also be equal.

That is by CPCT rule, corresponding parts of congruent triangles are equal.

Then,

$AB=PR$ ,

$BC=RQ$ ,

$AC=PQ$ ,

$\angle A=\angle P$ ,

$\angle B=\angle R$ and

$\angle C=\angle Q$ .

Thus,

$AB\ne PQ$ .
Therefore, the statement is false and

$0$ is the answer.

State whether the given statement is True(Press 1) or False(Press 0):
If the two angles and the included sides of one triangle are respectively equal to the two angles and the included side of the other triangle, then the two triangles are congruent.

If the two angles and the included sides of one triangle are respectively equal to the two angles and the included side of the other triangle, then the two triangles are congruent by the ASA rule.

Two right triangles are congruent, if hypotenuse and a side of one triangle are respectively equal to the hypotenuse and a side of the other triangle. If true enter 1 else 0.

A hypotenuse and one side of the right triangle is equal to the hypotenuse and ne side of another triangle. Thus, the two triangles will be congruent by RHS rule.

State true or false:
If the two sides of a triangle are unequal, then the larger side has the smaller angle opposite to it.
If true enter 1, else 0.

We know that if the two sides of a triangle are unequal, then the larger side has the largest angle opposite to it.

Hence, given statement is false.

Prove that the circle drawn with any equal side of an isosceles triangle as diameter bisects the base

$\displaystyle \bigtriangleup ABC$ in which AB=AC and A circle is drawn by taking AB as diameter which intersects the triangle at D join AD
IN

$\displaystyle \bigtriangleup ADB$ and

$\displaystyle \bigtriangleup ADC$ we are given AB=AC

$AD=AD$

$\displaystyle \angle ADB= 90^{\circ}$ (Angle in a semi-circle)

$\displaystyle \therefore \angle ADC = 90^{\circ}$

$\displaystyle \therefore \angle ADB = \angle ADC 90^{\circ}$
Hence

$\displaystyle \bigtriangleup ADB \cong \bigtriangleup ADC$
(RHS Theorem of congruence )

$BD=DC$ [ corresponding sides of congruent triangles]

AB is a line segment AX and BY are two equal line segments drawn opposite sides of line AB such that AX ||BY.
If line segments AB and XY intersect each other at point P. Prove that
(a) $\Delta APX \cong \Delta BPY$ (b) line segments AB and XY bisect each other at P.

In the given figure, segment

$AX$ and

$BY$ are equal and

$AX|| BY$

The line segment

$AX || BY$ and line segment

$AB$ join the parallel line at A and B.

And line segment

$XY$ cut line

$AB$ at point

$P$

Then, in

$\Delta AXP$ and

$\Delta PYB$

$AX=BY$ ....... (Given)

$\angle XAP(3)=\angle YBP(4)$ (Adjacent angle )

$\angle AXP(1)=\angle PYB(2)$ (Adjacent angle )

Then

$\Delta APX\cong \Delta BPY$

(B) The

$\Delta APX\cong \Delta BPY$

Then

$AP=BP$ and

$XP=PY$

Thus, the line segment

$AB$ and

$XY$ bisect each other.

In $\Delta ABC$ , the bisector AX of $\angle A$ intersects BC at X. $XL\perp AB$ and $XM \perp AC$ are drawn. Is XL = XM?
Why or why not?

Given :

$AX$ is the bisector of

$\angle A$ meet

$BC$ on

$X$

And

$XL\perp AB$ and

$XM\perp AC$

Then, in

$\Delta ALX$ and

$\Delta AMX$

$\angle LAX=\angle XAM$ ...... (

$AX$ is the bisector of angle A)

$\angle ALX=\angle AMX=90^{o}$ ...... (

$XL\perp AB$ and

$XM\perp AC$ )

$AX=AX$ ....... [Common side]

Then,

$\Delta ALX\simeq\Delta AMX$ ...... [By ASA rule of congruence]

$\therefore LX=MX$

Two parallel lines $l$ and $m$ are intersected by another pair of parallel lines $p$ and $q$ as in the figure. Show that triangles $ABC$ and $CDA$ are congruent.

Since

$l$ and

$m$ are parallel lines and AC is a transversal,

$\angle 1 = \angle 4$ . Similarly transversal AC cuts parallel lines

$p$ and

$q$ , so that

$\angle 2 = \angle 3$ .

In triangles ABC and CDA, we have

$\angle 1 = \angle 4$ (proved);

$\angle 2 = \angle 3$ (proved);

$AC = AC$ (common side).
By

$\text{A.S.A}$ postulate,

$\triangle ABC \cong \triangle CDA$ .

In the figure AD = BC and BD = CA. Prove that $\angle ADB = \angle BCA$ and $\angle DAB = \angle CBA$ .

In triangles

$ABD$ and

$BAC$ , we have

$AD = BC$ (given);

$AB = AB$ (common);

$AC = BD$ (given).
We can use SSS condition to conclude that

$\triangle ABD \cong \triangle BAC$ . From this we conclude that

$\angle ADB = \angle BCA$ and

$\angle DAB = \angle CBA$ .

In $\triangle ABC, AB = BC$ and $\angle B = 64^o$ . Find $\angle C$ .

We can see that

$\angle A$ =

$\angle C$ [

$\because AB= AC$ ]

$\angle A$ +

$\angle C$ +

$\angle B$ =

$180^\circ$ {

$\because$ sum of all the angles of triangle is

$180^o$ }
so,

$2\angle C$ =

$180^\circ$ -

$64^\circ$ =

$116^\circ$ (since

$\angle A$ =

$\angle C$ )

$\Rightarrow \angle C$ =

$\frac{116^\circ}{2}$ =

$58^\circ$

$\Rightarrow \angle A$ =

$58^\circ$ and

$\angle C$ =

$58^\circ$

$\Rightarrow \angle C$ =

$58^\circ$

In the figure, C is the mid point of AB, $\angle BAD = \angle CBE, \angle ECA= \angle DCB$ .
Prove that (a) $\Delta DAC \cong \Delta EBC$ (b) DA = EB.

(a) In given figure

$C$ is the mid point of

$AB$

$\angle BCE=\angle DCB+\angle ECD$ .....

$(i)$

$\angle DCA=\angle ECA+\angle ECD$ ......

$(ii)$

$\angle ECA=\angle DCB$ ..... (given )

$\therefore \angle BCE=\angle DCA$

Also,

$\angle BAD=\angle CBE$ ....... (Given)

and

$AC=BC$ ....... (Given)

$\therefore \Delta DAC\cong \Delta EBC$

(b) The

$\Delta DAC\cong \Delta EBC$

Then all corresponding angles and sides are equal

Then

$DA=EB$

In a quadrilateral ACBC, AC = AD and AB bisect $\angle A$ . Show that $\triangle ABC$ is congruent to $\triangle ABD$ .

In triangles

$ABC$ and

$ABD$ , we have

$AC = AD$ ........ (given);

$\angle CAB = \angle DAB$ .........

$(\because AB\ \text{ bisects }\ \angle A)$

$AB = AB$ ....... (common side).

Hence, by

$\text{S.A.S}$ criterion

$\triangle ABC \cong \triangle ABD$ .

$[hence \quad proved]$

In the figure, it is given that AE = AD and BD = CE. Prove that $\triangle AEB$ is congruent $\triangle ADC$ .

We have

$AE = AD$ and

$CE = BD$ .
Adding, we get

$AE + CE = AD + BD$

$\Rightarrow AC = AB$ .
In triangles

$AEB$ and

$ADC$ , we have

$AE = AD$ ....... (given);

$AB = AC$ ....... (proved);

$\angle EAB = \angle DAC$ , (common angle).
By SAS postulate

$\triangle AEB \cong \triangle ADC$ .

In $\triangle ABC, AD$ is the perpendicular bisector of $BC$ (see given figure). Show that $\triangle ABC$ is an isosceles triangle in which $AB = AC$

$AD$ is the perpendicular bisector

$\angle ADB=90^0 ---- (1)$ and

$BD=DC ---- (2)$

$\Delta ADB$ and

$\Delta ADC$

$\angle ADB= \angle ADC = 90^0$ from

$1$

$BD=DC$ From

$2$

$AD$ is the common side.

$\Delta ADB \cong \Delta ADC$ [by

$SAS$ congruence]

Thus sides

$AB=AC$

i.e.

$\Delta ABC$ is isosceles triangle.

Hence Proved

In the given figure, the point $P$ bisects $AB$ and $DC$ . Prove that
$\triangle APC\cong \triangle BPD$

In given figure P bisect AB and DC

Then P is the mid point of AB and CD

Then

$AP=PB$

And

$CP=PD$

Then in

$\Delta APC$ and

$\Delta BPD$

$AP=PB$

$CP=PD$

$\angle APC=\angle BPD$ ............(vertically opposite angles)

$\therefore \Delta APC\cong \Delta BPD$

The Indian Navy flight fly in a formation that can be viewed as two triangles with common side. Prove that $\triangle SRT \cong \triangle QRT$ , if $T$ is the midpoint of $SQ$ and $SR = RQ$ .

From the figure,

$RS$ is the common side and

$T$ is the mid point of

$SQ$ and also,

$SR=RQ$

$\bigtriangleup SRT and \bigtriangleup QRT$

$SR=RQ$ ....... (Given)

$ST=TQ$ ....... (

$T$ is the mid point of SQ)

$RT=RT$ ........ (Common side)

$\Rightarrow \triangle SRT \cong \triangle QRT$ ....... [By SSS condition of congruence]

$ABC$ is an isosceles triangle in which altitudes $BD$ and $CE$ are drawn to equal sides $AC$ and $AB$ respectively (see figure). Show that these altitudes are equal

ABC is an isosceles triangle in which altitudes BD and CE are drawn to equal sides AC and AB respectively .

Given ABC is an isosceles triangle

Then AB=AC

$\therefore\angle ABC=\angle ACB$

$\therefore\angle EBC=\angle DCB$

$\Delta EBC,$

$\Delta DCB$ .....................(1)

$\angle BEC=\angle CDB=90^{0}$

$BC=CB$

$\angle EBC =\angle DCB$

$\Delta BEC \cong \Delta CFB$ .......(SAA test)

$\therefore BD=EC$

Then these altitudes are equal.

In the figure given below, $AB = DC$ and $AC = DB$ . Is $\Delta ABC \cong \Delta DCB$ .

1348840_705152_ans_ca4b870664234cc8bacca6ec957fecfc.jpg

Complete the congruence statement.
$\Delta ABC \equiv$ ?

Solution:

Here

$AC=AT$

$\angle ABC=\angle ABT$

$BC=BT$

So,

$A\leftrightarrow A$

$B\leftrightarrow B$

$C\leftrightarrow T$

Hence, the congruence has proved

$\bigtriangleup ABC\equiv \bigtriangleup ABT$

DE = EC show that AB + BC > AD.

Sum of any two sides of any triangle > third side

$\implies\quad \triangle ABC$

$\implies\quad AB+BC>AC$

$\implies\quad AB+BC>AE+EC\quad -(1)\quad (\because AC=AE+EC)$

$In\triangle ADE$

$AE+DE>AD$

$\implies\quad AE+EC>AD\quad -(2)\quad (\because DE=EC)$

$Adding(1)\& (2)$

$\implies\quad AB+BC+AE+EC>AE+EC+AD$

$\implies\quad AB+BC>AD$

$hence\quad proved$

If $S$ is any point in the interior of $\Delta PQR$ , prove that $(SQ + SR) < (PQ + PR)$ .

Let's project

$QS$ to intresect

$PR$ at

$T$
From

$\triangle PQT$ , we have

$PQ+PT>QT$

$PQ+PT>SQ+ST$ ----------(1)

$ST+TR > SR$ ----------(2)

Adding equation(1) and (2),

$PQ+PT+ST+TR>SQ+ST+SR$

$PQ+PT+TR>SQ+SR$

$PQ+PR>SQ+SR$

$\Rightarrow SQ+SR<PQ+PR$

Hence proved.

983867_1028252_ans_79948e1f145b4265a917c7598bc433b9.png

Prove that a median divides a triangle into equal parts

$In\triangle ABC,AD$ is median

$BD=DC$

$draw\quad AE\bot BC$

area of

$\triangle ABD=\cfrac { 1 }{ 2 } \times BD\times AE\quad [since\quad BD=DC]$

=area of

$\triangle ADC$

$\therefore$ Median of a

$\triangle gle$ divides it into two

$\triangle gles$ of equal area.

1013031_962111_ans_87b71237290348a39dcf2afb1562288e.png

In the adjoining figure $\angle{ABD}={132}^{o}$ & $\angle{EAC}={120}^{o}$ . Prove that $AB>AC$

$\angle ABC+\angle ABC=180^{\circ}$

$\angle ABC=180-132=48^{\circ}$

$\angle EAC+\angle BAC=180^{\circ}$

$\angle BAC=180^{\circ}-120^{\circ}=60^{\circ}$

$\angle ABC+\angle BCA+\angle CAB=180^{\circ}$

$\angle BCA=180-60-48=72^{\circ}$

$\angle BCA$ is largest angle.

$\angle ABC$ is smallest angle

$\Rightarrow$ sides opposite to largest angle are large (AB)

sides opposite to smallest angle are small (AC)

$\therefore ABC > AC$

Write down the total number of True statement:
Take any point O in the interior of a triangle PQR.
Is
(i) OP + OQ > PQ?
(ii) OQ + OR > QR?
(iii) OR + OP > RP?

$In \ \triangle OPQ$

i) From property of triangles,

$OP+QO>PQ$

(

$\because$ Sum of two sides > third side)

ii)

$\triangle OQR:$

$OQ+QR>OR$

iii) Again, using same property in

$\triangle POR$

$OP+OR>PR$

$\therefore$ All 3 statements are true.

"If two sides and an angle of one triangle are equal to two sides and an angle of another triangle, then the two triangles must be congruent". Is the statement true? Why?

The statement is false.
If two sides and the included angle of one triangle are equal to corresponding two sides and the included angle of another triangle then the two triangles must be congruent.

In figure The sides $BA$ and $CA$ have been produced such that $BA=AD$ and $CA=AE$ prove that segment $DE\parallel BC$

$\Delta ABC$ and

$\Delta ADE$

$BA=AD$

$CA=AE$

$\angle BAC=\angle EAD$ (vertically opposite angles)

$\Delta ADA\cong \Delta BAC$ (SAS)

$ED||BC$ (cpct)

1077020_1044629_ans_11faf14b6674484d9379450d81ebfcaa.png

In triangles $ABC$ and $PQR, \angle A = \angle Q$ and $\angle B = \angle R$ . Which side of $\Delta PQR$ should be equal to side $AB$ of $\Delta ABC$ so that the two triangle are congruent? Give reason for your answer.

According to given condition,

For congruency of triangle all sides and angle of one triangle should be equal to other triangle's corresponding parameters.

So,

$QR=AB$

In fig., PQRS is square and SRT is an equilateral triangle. Prove that
(i) $PT = QT$ (ii) $\angle TQR = {15^ \circ }$

(i) Because PQRS is a square

$\angle PSR = \angle QRS = 90^{ \circ}$

Now In $\triangle SRT$

$\angle TSR = \angle TRS = 60^{ \circ}$

$\angle PSR + \angle TSR = \angle QRS + \angle TRS$
$\implies \angle TSP = \angle TRQ$
Now in $\triangle TSP$ and $\triangle TRQ$

$TS = TR$

$\angle TSP = \angle TRQ$

$PS = QR$

Therefore , $\triangle TSP ≡ \triangle TRQ$

So $PT = QT$

(ii) Now in $\triangle TQR$ ,

$TR = QR ( RQ = SR = TR )$

$\angle TQR = \angle QTR$

And $\angle TQR + \angle QTR + \angle TRQ = 180$

$\implies \angle TQR + \angle QTR +\angle TRS + \angle SRQ = 180$

$\implies 2( \angle TQR) + 60 + 90 = 180$ $( \angle TQR = \angle QTR)$

$2 ( \angle TQR ) = 30$

$\implies \angle TQR = 15^{ \circ}$

Give any two real-life examples for congruent shapes.

$\Rightarrow$ If two figures are congruent, then they're exactly the same shape, and they're exactly the same size.

$\Rightarrow$ They may appear different because one is shifted or rotated a certain way, but they're still the same shape, and all the sides of one are the same length as the corresponding sides of the other.

$\Rightarrow$ Two real-life examples for congruent shapes :

$(1)$ Biscuits of the same pack.

$(2)$ Earrings of the same set.

In figure, $PR>PO$ and $PS$ bisect $OPR$ . Prove that $\angle{PSR}>\angle{PSO}$

Given:

$PR > PO$ and

$PS$ bisects

$\angle OPR$

To prove:

$\angle PSR > \angle PSO$

Proof:

$\angle POR > \angle PRO$

$..........(1)$

$(PR > PO$ as angle opposite to larger side is larger.)

$\angle OPS = \angle RPS$

$............(2)$

$(PS$ bisects

$\angle OPR$ )

$\angle PSR = \angle POR +\angle OPS$

$...........(3)$

(exterior angle of a triangle equals to the sum of opposite interior angles)

$\angle PSO = \angle PRO + \angle RPS$

$........(4)$

(exterior angle of a triangle equals to the sum of opposite interior angles)

Adding

$(1)$ and

$(2)$

$\angle POR + \angle OPS > \angle PRO + \angle RPS$

$\Rightarrow \angle PSR > \angle PSO$

$[$ from

$(1), (2), (3)$ and

$(4)]$

Hence, this is the answer.

If the point $p\left( {{a^2},\,a} \right)\,$ lies in the region. Corresponding in the acute angle between the lines $2y = x\,\,and\,\,4y = x,$ then find the value of $a$ or the range in which $a$ lies.

$x = 2y,4y = x$

$4 = x - 2y,v = x - 4y$

Joint eq’s

$u.v = 0$

$\left( {x - 2y} \right)\left( {x - 4y} \right) = 0$

${x^2} - 6yx + 8{y^2} = 0$

$s\left( {x,y} \right) = {x^2} - 6xy + 8{y^2} = 0$

If point $p\left( {{a^2},a} \right)$ , lies in interior of awte angle formed by these lines,

Then $s\left( {x,y} \right)$ at $p\left( {{a^2},a} \right) < 0$

$\Rightarrow {\left( {{a^2}} \right)^2} - + {a^2} - a + 8{a^2} < 0$

$\Rightarrow {a^2}\left( {{a^2} - 6a + 8} \right) < 0$

$\Rightarrow {a^2} - 6a + 8 < 0$

$\Rightarrow \left( {a - 2} \right)\left( {a - 4} \right) < 0$

$a \in \left( {2,4} \right)$

In the given fig $\triangle PQC$ & $PRC$ such that $QC=CR$ , $PQ=PR$ . Prove that $\triangle PQC\cong\triangle CPR$

$PQ=PR$ (Given)

$\therefore \angle PQR=PRQ$ (Angles opposite to equal sides)

$QC=RC$ (Given)

$\triangle PQC\cong \triangle CPR$

$SAS$ criteria

If the area of two similar triangles are equal, prove that they are congruent.

Given that:

$ar \triangle ABC =ar \triangle DEF$

Prove that:

$\triangle ABC\cong \triangle DEF$

Proof:

Two triangles ABC and DEF are shown in figure.

$ar \triangle ABC =ar \triangle DEF$ (given)

$\dfrac{ar\triangle ABC} {ar \triangle DEF} = 1$

$\dfrac{AB^2}{DE^2} = \dfrac{BC^2}{EF^2} = \dfrac{CA^2}{FD^2}= 1$ [ USING THEOREM OF AREA OF SIMILAR TRIANGLES]

$AB= DE , BC= EF, CA= FD$

Thus,

$\triangle ABC\cong \triangle DEF$ [BY SSS criterion of congruence]

997029_1051413_ans_daed264863914f56892b2b962ffadf47.jpg

Give any two real-life examples of congruent shapes.

$\Rightarrow$ Two polygons are congruent if they are the same size and shape - that is, if their corresponding angles and sides are equal.

$\Rightarrow$ Two real-life examples of congruent shapes are :

$(1)$ Pages of book.

$(2)$ Two same mobile phones.

$S$ and $T$ are respectively the mid points of equal sides $PQ$ and $PR$ of $\triangle {PQR}$ . Show that $Qt=RS$ .

$\triangle PQR$ ,

$PQ=PR$ ------ [ Given ]

$\triangle PQR$ is an isosceles triangle.

$\therefore$

$\angle PQR=\angle PRQ$ [ Angles of isosceles triangle ]

i.e

$\angle SQR=\angle TRQ$ ------ ( 1 )

Here,

$S$ and

$T$ are mid-point of

$PQ$ and

$PR$

$\therefore$

$SQ=TR$ ----- ( 2 )

Now, in

$\triangle SQR$ and

$\triangle TRQ$

$\Rightarrow$

$SQ=TR$ [ From ( 2 ) ]

$\Rightarrow$

$\angle SQR=\angle TRQ$ [ From ( 1 ) ]

$\Rightarrow$

$QR=RQ$ [ Common side ]

$\therefore$

$\triangle SQR\cong\triangle TRQ$ [ By

$SAS$ criteria ]

$\therefore$

$RS=QT$ [ By c.p.c.t ]

993353_1060667_ans_6fd00380143048d8bc280b2f9ce9d3ba.jpeg

You want to show that $\Delta ART \cong \Delta PEN$ .
If it is given that $AT = PN$ and you are to use ASA criterion, you need to have
i) ?
ii) ?

According to

$ASA$ criterion , If two angles and their included side in one triangle are equal to the two angles and their included side in other triangle, then the triangles are congruent.

So, if side

$AT=PN$ in the given two triangles.

The triangles are congruent if,

$\angle RAT=\angle EPN$

$\angle ATR=\angle PNE$

$\Delta$ ABC and $\Delta$ DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC. If AD is extended to intersect BC at P, show that $\Delta$ ABP $\cong$ $\Delta$ ACP.

To prove :

$\triangle ABP\cong \triangle ACP$

Proof: In

$\triangle ABD$ and

$\triangle ACD$

$AB=AC$ [Given]

$BD=DC$ [Given]

$AD$ is common

$\triangle ABD \cong \triangle ACD$ [SSS postulate]

$\hat { BAD }= \hat { CAD }$ [CPCT]

$\Rightarrow \hat { BAP } = \hat {CAP }$

$\triangle ABP$ and

$\triangle ACP,$

$AB=AC$ [Given]

$AP$ is common\]

$\hat { BAP } = \hat { CAP }$ [proved]

$\therefore \triangle BAP\cong \triangle CAP$ [By SAS postulates]

1331596_1072867_ans_b7e8386101ca4863b8ecf716e09ae542.png

In $\triangle PQR, PQ=PR$ and $m \angle P=40^{o}$ then $m \angle Q=.$

Given that

$PQ=PR$ then

$\angle R=\angle Q=x$

$\angle P+\angle Q+\angle R=180^\circ$

$x+x+40^\circ=180^\circ$

$x=70^\circ.$

$\angle R=\angle Q=70^\circ.$

1056967_1085842_ans_70c206e9252d4b1f84bc50ebbba28c49.jpg

In a $\triangle ABC,BC=a,CA=b$ and $\angle BCA=120^{o},CD$ is angle bisector of $\angle BCA$ which meets $AB$ on $D$ find the length of $CD$

Given In

$\triangle ABC, BC=a,CA=b$ and

$\angle BCA=\angle C=120^{\circ}$

as CD is angle bisector of

$\angle C$

$\implies$ length of

$CD= \dfrac{2{a}{b}}{a+b}\cos{\dfrac{C}{2}} =\dfrac{2{a}{b}}{a+b}\cos60^{\circ}=\dfrac{{a}{b}}{a+b}$

In the given figure $\Delta ABC \cong \, \Delta ABT$ , write all the corresponding sides.

From the given figure, we can say that,

$AC\cong AT$

$BC\cong BT$

$AB\perp CT$

In the adjoining figure, PQ $=$ PR and QL $=$ MR. Prove that PL $=$ PM.

$\triangle PQL$ and

$\triangle PRM$

$PQ=PR$ [ Given ]

$\angle PQL=\angle PRM$ [ angles opposite to equal sides are also equal ]

$QL=MR$ [ Given ]

$\therefore$

$\triangle PQL\cong\triangle PRM$ [ By SAS congruence rule ]

$\Rightarrow$

$PL=PM$ [ CPCT ]

1258700_1080537_ans_bdd018ab6dcc4369886c4110146295b4.png

Explain, why ?
$\Delta A B C = \Delta F E D$

1046113_1179125_ans_f95fc106133e43f890e5d301951ee90b.png

In figure $ABCD$ , $ABFE$ and $CDEF$ are parallelograms. Prove that $ar(\triangle {ADE})=ar(\triangle{BCF)}$

$\Delta ADE$ &

$\Delta BCF \rightarrow$

$AD=BC\,(ABCD$ is a parallelogram) (opp. sides of

$11 gm)$

$AE=BE\,(ABFE$ is a parallelogram) (opp. sides of

$11 gm)$

$ED=FC\, (CDEF$ is a

$11 gm)$ (opp. sides of

$11 gm)$

$\therefore \Delta ADE \cong \Delta BCF$ (by SSS)

Since

$\Delta ADE$ &

$\Delta BCF$ are congruent

$arc(\Delta ADE)= arc (\Delta BCF)$

1054619_1158063_ans_a19b721255e449fea64aab4fec5b2500.png

If $\bigtriangleup ABC\cong\bigtriangleup FED$ $under\ the\ corresponding \ ABC \leftrightarrow FED.$ Write all the corresponding congruent part of the triangle.

Given:-

$\triangle{ABC} \cong \triangle{FED}$

Here,

$A \leftrightarrow F$

$B \leftrightarrow E$

$C \leftrightarrow D$

Therefore, by

$C.P.C.T$ ., i.e., corresponding parts of congruent triangles,

Corresponding angles are equal, i.e.,

$\angle{A} = \angle{F}$

$\angle{B} = \angle{E}$

$\angle{C} = \angle{D}$

Corresponding sides are equal, i.e.,

$AB = FE$

$BC = ED$

$AC = FD$

In the following figure, $\angle A = \angle C$ and $AB = BC.$
Prove that $\Delta ABD \cong \Delta CBE$

In triangle

$ABD$ and

$CBE$ ,

$AB=BC$ ( given )

$\angle BAD=\angle BCE$ ( given )

$\angle ABD=\angle CBE$ ( common angle )

By ASA criteria of congruence,

$\triangle ABD\cong \triangle CBE$

Prove that in a right angled triangle, hypotenuse is the longest side.

Let’s name triangle $ABC$

In triangle $ABC$

Angle $B={{90}^{0}}$

Angle B and angle $A,$

$AC>BC$ ………..(1) (side opposite to greater angle is longer)

Angle $B$ and angle $c,$

$AC>AB$ ……..(2) (side opposite to greater angle is longer)

From equation 1^st and 2^nd ,

$AC>BC,AB$

Hence, hypotenuse is the longest side.

Given: $\Delta A \bot AB$ , $CB \bot AB$ and $AC = BD$
State $3$ points of equal pairs in $\Delta ABC$ & $\Delta ABD$ & Congruence rule.

$\triangle ABC$ and

$\triangle ABD$

$\angle A=\angle B=90^{\circ}$ [A]

$AC=BD$

$\Rightarrow AD=BC$ [S]

$AD\parallel BC$ [S]

From SAS theorem

$\triangle ABC \cong \triangle ABD$

ABC is a right isosceles triangle right angled at A. Find the value of $\angle B$ and $\angle C$ .

$\angle A =90^{o}$

$\angle B =\angle C$ (Given the triangle is isosceles)

$\triangle ABC$

$\angle A +\angle B +\angle C =180^{o}$

$90 +\angle B +\angle C =180^{o}$

$\angle B +\angle C =90^{o}$

$\angle B +\angle B =90^{o}$

$2\angle B =90^{o}$

Hence,

$\angle B =\angle C =45^{o}$ .

In the given figure, equal sides $BA$ and $CA$ of $\Delta ABC$ are produced to $Q$ and $P$ respectively such that $AP = AQ$ . Prove that $PB = QC$ .

$\Delta APB \cong \Delta AQC$

PA is tangent to the circle with centre O. If BC = 3cm, AC = 4 cm and $\Delta ACB \sim \Delta PAO,$ then find OA and $\frac{OP}{AP}$ .

Consider the problem

$\angle ACB$ is an angle in a semi circle

Therefore,

$\angle ACB$

$= {90^ \circ }$

Now, in

$\Delta ACB$

we use Pythagoras Theorem

${\left( {AB} \right)^2} = {\left( {AC} \right)^2} + {\left( {BC} \right)^2}$

$\begin{array}{l} \therefore \, A{ B^{ 2 } }=16+9 \\ AB=\sqrt { 25 } \\ =5cm \end{array}$

So,

Radius

$OA=\dfrac { 1 }{ 2 } AB$

$\begin{array}{l} OA=\dfrac { 1 }{ 2 } \times 5 \\ =\dfrac { 5 }{ 2 } cm \end{array}$

Now, since

$\Delta ACB \sim \Delta PAO$

$\dfrac{{BA}}{{CA}} = \dfrac{{OP}}{{AP}}$

Therefore,

$\,\dfrac{{OP}}{{AP}} = \dfrac{5}{4}$

Is it possible to have a triangle with the following sides?
2 cm, 3 cm,5 cm

Sum of any two sides of triangle must be strictly greater than third side ,

But here

$2+3=5$

So it is not possible to construct such triangle

In figure, equal sides have been marked by the same signs.
Can we say $\Delta ABC\cong \Delta DBC$ ?

$\triangle ABC$ and

$\triangle DCB$

$BC=BC$ (common)

$AB=DC$ (given)

$AC=BC$ (given)

$\therefore \triangle ABC \cong \triangle DCB$ [by SSS]

1198096_1215250_ans_cda67213d43747048d26886576bfebe6.png

In $\triangle ABC$ , AD is perpendicular bisector of BC (See adjacent figure). Show that $\triangle ABC$ is an isosceles triangle in which $AB=AC$

Given:

$AD$ is the perpendicular bisector of

$BC$ .

To Prove:

$\triangle{ABC}$ is an isosceles triangle. i.e,

$AB = AC$

Proof: In

$\triangle{ADB}$ and $$\triangle{ADC},

$AD = AD$ (Common)

$\triangle{ADB}=\triangle{ADC}$ . ( each

${90}^{\circ}$ )

$\Rightarrow BD=CD$

where

$AD$ is the perpendicular bisector

Therefore,

$\triangle{ADB}\cong \triangle{ADC}$ ( by SAS congruence rule)

$\Rightarrow AB=AC$ (by CPCT)

So,

$\triangle{ABC}$ is an isosceles triangle.

Note: In Congruent Triangles corresponding parts are always equal and we write it in short CPCT i e, corresponding parts of Congruent Triangles.

In the figure, $AP =AQ$ and $BP= BQ.$

Prove that $AB$ is the bisector of $\angle PAQ$ and $\angle PBQ.$

$\triangle APB$ and

$\triangle AQB$

$AP=AQ$ (Given)

$BP=BQ$ (Given)

$AB=AB$ (Common Side)

$\therefore \triangle APB\cong\triangle AQB$ (S.S.S. congruency)

$\Rightarrow \angle PAB=\angle QAB$ (CPCT)

$\Rightarrow \angle PBA=\angle QBA$ (CPCT)

Thus, AB is the bisector of

$\angle PAQ$ and

$\angle PBQ.$

In figure, $AB=AC$ and $\angle 1=\angle 2$ . Is $\Delta ABD\cong \Delta ACD$ ?(Give reason)

1114635_1203432_ans_fa30a7fcacd8447ea836b294cee74c5b.jpg

If $\Delta A B C \cong \Delta F E D$ under the correspondence $A B C \leftrightarrow F E D$ , write all the corresponding congruent parts of the triangles.

Given that triangle ABC is Congruent to triangle FED. The corresponding congruent parts of the triangle are:

$\angle A\leftrightarrow \angle F$

$\angle B \leftrightarrow \angle E$

$\angle C \leftrightarrow \angle D$

$AB \leftrightarrow FE$

$BC \leftrightarrow ED$

$AC \leftrightarrow FD$

1191029_1212383_ans_b84ef79429fd477a8785dc3ea9f8b4fe.PNG

If $\Delta A B C \cong \triangle N M O ,$ name the congruent sides and angles.

$\triangle ABC \cong \triangle NMO,$ then

Congruent Sides:

$AB=NM,BC=MO,AC=NO$

Congruent Angles:

$\angle ABC=\angle NMO,\angle BCO=\angle MON,\angle BAC=\angle MNO$

Consider the figure, $PQRS$ is a rectangle. $PT$ and $RU$ are perpendicular from $P$ and $R$ on $SQ$ . Prove that $\Delta PTS\cong \Delta QRU$

REF.Image.

$\triangle PTS$ and

$\triangle QUR$

$\angle PTS = \angle QUR = 90^{\circ}$ (given)

$\angle TSP = \angle UQR$ [alternate angles ]

$\Rightarrow \angle TPS = \angle QRU$

$SP = QR$ (given)

$\therefore$ By Angle side Angle (A.S.A) axiom of congruence.

$\triangle PTS \cong \triangle RUQ$

1119809_1203468_ans_9f7935e43b684ff5ac10d4a855a4de7f.png

In figure, equal sides have been marked by the same signs.
Is $\Delta ABC\cong \Delta DCB$ ?

$\triangle ABC$ and

$\triangle DCB$

$BC=BC$ (common)

$AB=DC$ (given)

$\therefore \triangle ABC \cong \triangle DCB$ [by SAS]

1198070_1215237_ans_14a63267f25e4f33a3439cde6fd923e4.png

In the figure, $\Delta CDE$ is an equilateral triangle formed on a side $CD$ of a square $ABCD$ . Show that $\Delta ADE \cong \Delta BCE$ .

(Refer to Image)

Given

$ABCD$ is a square.

$CDE$ is an equilateral

$\triangle$ .

To prove that

$\Delta ADE \cong \Delta BCE$

Proof:

$AD=BC$ (sides of a square)

$DE=EC$ (sides of equilateral

$\triangle$ )

$\angle ADE=\angle BCE=(90^o+60^o)=150^o$

Hence,

$\Delta ADE \cong \Delta BCE$

$SAS$ congruency.

1207825_1243918_ans_56bd38c4385c4c5489263a6712bdf291.JPG

In the given figure, prove that:
$AC=DB$

$AB=CD$ (given)

$\angle DCB =\angle ABC = 90^o$

$BC = BC$ (common)

By SAS congruence

$\triangle ABC \cong \triangle DCB$

so,

$AC = DB$ (corresponding sides of congruent triangles)

One of the angles of a $\triangle$ is ${ 75 }^{ }.$ If the difference of the other two other is ${ 35 }^{ }.$
Find the largest angle of the $\triangle .$

We know that,

$A+B+C=180^0$

Now,

$\begin{array}{l} b-c=35 \\ 75+b+c=180 \\ b+c=105 \\ b-c=35 \\ solving\, \, both\, \, \, sides, \\ 2b=105+35 \\ b=\dfrac { { 140 } }{ 2 } \\ \therefore b=70 \\ b-c=35 \\ 70-c=35 \\ c=35 \\ \therefore a=75\, \, \, b=70\, \, \, c=35 \end{array}$

Hence,

We get the largest angle

$a=75^0$

In $\triangle{ABC}$ , $AB=AC$ and $AD$ is the perpendicular bisector of $BC$ . Show that $\triangle{ADB}\cong\triangle{ADC}$

R.E.F image

Given :

$AD$ is the perpendicular

bisector of

$BC ,AB=AC$

To show

$\bigtriangleup ADB\cong \bigtriangleup ADC$

proof :- In

$\bigtriangleup ADB$ and

$\bigtriangleup ADC$

$AD=AD$ (common)

$\angle ADB=\angle ADC$

$BD=CD$ (

$AD$ is perpendicular bisector)

$\therefore \bigtriangleup ADB \cong \bigtriangleup$ ADC by SAS congruence .

In the given figure, prove that:
$\triangle ABC\cong \triangle DCB$

$AB=CD$ (given)

$\angle DCB =\angle ABC$

$BC = BC$ (common)

By SAS congruence

$\triangle ABC\cong \triangle DCB$

Hence Proved

If the area of two similar triangles are equal, prove that they are congruent.

Given:-

$ar{\left( \triangle{ABC} \right)} = ar{\left( \triangle{DEF} \right)} \\ \triangle{ABC} \sim \triangle{DEF}$

To prove

$\triangle{ABC} \cong \triangle{DEF}$

Proof:-

$\because \triangle{ABC} \sim \triangle{PQR} \quad \left( \text{Given} \right)$

$\therefore \cfrac{{AB}^{2}}{{DE}^{2}} = \cfrac{{BC}^{2}}{{EF}^{2}} = \cfrac{{CA}^{2}}{{DF}^{2}} = \cfrac{ar{\left( \triangle{ABC} \right)}}{ar{\left( \triangle{DEF} \right)}}$

$\because ar{\left( \triangle{ABC} \right)} = ar{\left( \triangle{DEF} \right)}$

$\therefore \cfrac{{AB}^{2}}{{DE}^{2}} = 1$

$\Rightarrow AB = DE$

Similarly,

$BC = EF$

$AC = DF$

Now,

$\triangle{ABC}$ and

$\triangle{DEF}$ ,

$AB = DE$

$BC = EF$

$AC = DF$

By SSS congruency,

$\triangle{ABC} \cong \triangle{DEF}$

Hence proved.

1350522_1339191_ans_60c58e8e9fee491ea58b9ec03bb6ded4.png

I the given figure, we have $PQ=SR$ and $PR=SQ$ .
Prove that $\triangle PQR \cong \triangle SRQ$ .

$\triangle{PQR}$ and

$\triangle{SRQ}$ ,

$PQ = SR \quad \left( \text{Given} \right)$

$QR = QR \quad \left( \text{Common} \right)$

$PR = SQ \quad \left( \text{Given} \right)$

$SSS$ property,

$\triangle{PQR} \cong \triangle{SRQ}$

Hence proved.

In the given $AB$ and $CD$ bisect each other at $O$ .State the three pairs of equal parts in two triangles $AOC$ and $BOD$

From the given figure,

$OC = OD$

$OA = OB$

$\angle{COA} = 180^o-\angle COB ..... (1)$ (linear pair)

$\angle{DOB} = 180^o-\angle COB....... (2)$ (linear pair)

$\angle{COA} = \angle{DOB}$ from

$(1) \& (2)$

If $\Delta PQR$ is an isosceles triangle such that PQ=PR , then prove that the attitude PS from P on QR bisects QR

We have,

According to given figure.

$PQ=PR\,\,\,\left( given\,\,that \right)$

$QS=SR\,\,\,\left( By\,defination\,of\,mid\,po\operatorname{int} \right)$

$PS=PS\,\,\,\left( Common\,line \right)$

Then,

$\Delta SPQ\cong \Delta SPR$ (BY congruency S.S.S.)

Hence, PS bisects

$\angle \,PQR$ by definition of angle bisector.

You want to establish $\triangle DEF\cong \triangle MNP$ , using ASA congruence rule. You are given that $\angle D=\angle M$ and $\angle F=\angle P$ . What information is needed to establish the congruence ?

Given:-

$\angle{D} = \angle{M} \\ \angle{F} = \angle{P} \\ \triangle{DEF} \cong \triangle{MNP} \quad \left( \text{By ASA congruent rule} \right)$

As we know that if angle on either side are equal then the sides are also equal.

Thus the

$DF = MP$ is the needed information to establish the congruence.

In pair of triangle in the following figure, parts bearing identical marks are congruent. State the test and correspondence if vertices by which triangle in each pair are congurent.

Consider the triangles

$\Delta XWZ$ and

$\Delta YXZ$

Here

$WX=WY$ (Given )

$WZ=WZ$ (common side )

$\angle XWZ=\angle YWZ$ (Given)

So, by

$SAS$ criteria

The trriangles are congruent

Check whether $\triangle{ABC}\cong\triangle{DEF}$ .Give Reason
In $\triangle ABC: AB=6\ cm,\angle B=50^{\circ}$ and $\angle C=90^{\circ}$
In $\triangle DEF: DE=6\ cm,\angle E=50^{\circ}, \angle F=90^{\circ}$ .

REF.Image

As sum of angles in a triangle is

$180^{\circ}$

$\Rightarrow \angle A= 40^{\circ}, \angle D= 40^{\circ}$

$\Rightarrow \angle A=\angle D, AB= DE, \angle B=\angle E$

$\Rightarrow$ By ASA congruent, triangles are congruent

1194230_1365802_ans_82d64c8a6b4d4cea96e157ba8301ec73.jpg

In the given figure, $\overline{PL}\bot \overline{OB}$ and $\overline{PM}\bot \overline{OA}$ such that $\overline{PL}=\overline{PM}$ . Is $\triangle PLO\cong \triangle PMO$ ? Give reasons in support of your answer.

Consider

$\Delta PLO \>and \> \Delta PMO$

$\angle L=\angle M=90\\PM=PL\\PO=PO$

So by

$RHS$ criteria

The triangles are congruent

Write converse of the theorem "In $\triangle ABC$ , if $AB=AC$ then $\angle C=\angle B$ "

Converse states that "The sides opposite to equal angles of a triangle are equal".

As AB=AC

$\rightarrow$

$\angle{C}=\angle{B}$

In the given figure, PS is the bisector of $\angle QPR$ of $\Delta PQR$ Prove that $\frac{QS}{SR}=\frac{PQ}{PR}$

$\Delta QSP \ and \ \Delta RSP$ ,

$QS=SR , SP=SP$

Then by SS similarity criterion,

$\Delta QSP\sim \Delta RSP$

Therefore,

$\dfrac{QS}{SR}=\dfrac{QP}{RP}$ (In similar triangles corresponding sides are proportional)

Is it possible to construct a triangle with sides 9 cm, 6 cm and 17 cm ? If not, why ?

It is not possible to construct a triangle with sides 9 cm, 6 cm and 17 cm because 9+6 < 17 .

Take any point $O$ in the interior of a $\triangle{PQR}$ .Is $OR+OP>RP$ ?

$\triangle{ORP}$ ,

Sum of any two sides is greater than the third side

$\therefore OR+OP>RP$

So, this is true.

State SSS congruency Rule of traingles.

SSS Criterion (Side - Side -Side Criterion) :

If all three sides of one triangle are equal to the corresponding sides of another triangle, then the two triangles are said to be congruent by SSS Rule.

In the given figure,

$AB=PQ,QR=BC$ and

$AC=PR$

Hence,

$\triangle{ABC}\cong\triangle{PQR}$

1264726_1396134_ans_7d1057f971214b55ae18ac8f45326cb2.PNG

Take any point O in the interior of a triangle PQR. is
$OQ+OR>QR?$

$\triangle{OQR}$ ,

Sum of any two sides is greater than the third side

$\therefore OQ+OR>QR$

So, this is true.

As shown in the figure, Avinash is standing near his house. He can choose from two roads to go to school. Which way is shorter? Explain why.

1326002_1565938_ans_548d11f295a04f4f950d67174be11ca6.jpeg

Give any two real time examples for congruent shapes.

The two real life examples for congruent shapes are:

$(i)$ Pair of ear- rings.

$(ii)$ Pages of same book

Is $\Delta ABC\cong \Delta DEF?$ Give reasons in support of your answer.

REF.Image

$\angle B=\angle E$ ......(given)

$\angle A=\angle D$ .......(given)

side

$BC=$ side

$EF$ ......(given)

$\therefore$ By ASA (Angle Side Angle) test of similarity

$\Delta ABC \cong \Delta DEF$

1210386_1507579_ans_7df449a9b06a4f06a532d323ed529989.png

In the following, pairs of triangles of Fig., the lengths of the sides are indicated along the sides. By applying SSS congruence criterion, determine which triangles are congruent. If congruent, write the results in symbolic form.

In triangle

$AOB$ and

$DOE$ ,

Given,

$AO = DO = 2cm$

$DE = AE = 2cm$

$EO = BO = 1.5cm$

$\therefore$

$\triangle AOB \cong \triangle DOE$ .. SSS criterion

Without drawing the triangles write all size pairs of equal measures in each of the following pairs of congruent triangles.
$\triangle ABC \cong \triangle LMN$

$\triangle ABC \cong \triangle LMN$

$AB=LM, BC=MN, AC=LN$

$\angle ABC=\angle LMN, \angle ACB=\angle ANM, \angle BAC=\angle MLN$

In the following, pairs of triangles of Fig., the lengths of the sides are indicated along the sides. By applying SSS congruence criterion, determine which triangles are congruent. If congruent, write the results in symbolic form.

In triangle

$LMN$ and

$LON$ ,

$LM = LO = 5cm$

$MN = ON = 5.5cm$

$LN = LN$ [Common]

$\therefore$

$\triangle LMN \cong \triangle LON$ ...SSS criterion

In $\triangle PQR$ and $\triangle SQR$ are both isosceles triangle on a common base $QR$ such that $P$ and $S$ lie on the same side of $QR$ . Are triangles $PSQ$ and $PSR$ congruent ? Which condition do you use ?

Let triangle SQR and PQR are the given triangles such that

$PR = PQ$ and

$SR = SQ$

In triangle

$PSR$ and

$PSQ$ ,

$PR = PQ$

$SR = SQ$

$PS = PS$

$\therefore$

$\triangle PSR \cong \triangle PSQ$ ...SSS criterion

Yes, both triangles

$PSR$ and

$PSQ$ are congruent with the SSS criterion.

In Fig.6.49, it is given that $LM = ON$ and $NL = MO$
(a) State the three pairs of equal parts in the triangles $NOM$ and $MLN$ .
(b) Is $\triangle NOM \cong \triangle MLN$ . Give reason ?

(a) In triangle

$MLN$ and

$NOM$

$NM = MN$

$LN = OM$

$LM = ON$

(b) By using part (a) of the solution, we get

$\therefore$

$\triangle MLN \cong \triangle NOM$ .. SSS criterion

In the following, pairs of triangles of Fig., the lengths of the sides are indicated along the sides. By applying SSS congruence criterion, determine which triangles are congruent. If congruent, write the results in symbolic form.

In triangle

$XYZ$ and

$YXW$ ,

Given,

$XY = YX$ [Common]

$ZY = WY = 3\ cm$

$ZX = WY = 5\ cm$

$\therefore$

$\triangle XYZ \cong \triangle YXW$ ...SSS criterion

In the following, pairs of triangles of Fig. the lengths of the sides are indicated along the sides. By applying SSS congruence criterion, determine which triangles are congruent. If congruent, write the results in symbolic form.

In triangle

$STU$ ,

By Pythagoras Theorem,

$\Rightarrow$

$(TU)^{2} = (10.5)^{2} - 5^{2}$

$\Rightarrow$

$(TU)^{2} = 135.25$

Now, in triangle

$PQR$ ,

$\Rightarrow$

$(QR)^{2} = (10.5)^{2} - 5^{2}$

$\Rightarrow$

$(QR)^{2} = 135.25$

$\therefore$

$(QR)^{2} = (TU)^{2}$

$\Rightarrow$

$(QR) = (TU)$ ...(1)

In triangle

$STU$ and

$POR$ ,

$ST = PQ = 5cm$

$SU = PR = 10.5cm$

$QR = TU$ from 1

$\therefore$

$\triangle STU \cong \triangle PQR$ ...SSS criterion

In the following, pairs of triangles of Fig. the lengths of the sides are indicated along the sides. By applying SSS congruence criterion, determine which triangles are congruent. If congruent, write the results in symbolic form.

In triangle

$PQR$ and

$RSP$ ,

Given,

$PR = RP$

$PQ = RS = 5cm$

$OR = SP = 3cm$

$\therefore$

$\triangle PQR \cong \triangle RSP$ .. SSS criterion

In the following, pairs of triangles of Fig. the lengths of the sides are indicated along the sides. By applying SSS congruence criterion, determine which triangles are congruent. If congruent, write the results in symbolic form.

In triangle

$PQR$ and

$RSP$ ,

Given,

$SU = SU$

$ST = SV = 5cm$

$UT = UV = 3cm$

$\therefore$

$\triangle SVU \cong \triangle STU$ .... SSS criterion

Without drawing the triangles write all size pairs of equal measures in each of the following pairs of congruent triangles.
$\triangle STU \cong \triangle DEF$

$\triangle STU \cong \triangle DEF$

$ST=DE, TU=LF, SU=DF$

$\angle STU=\angle DEF, \angle TSU=\angle EDF, \angle SUT=\angle DFE$

check if the following pair of triangles is congruent? also, state the condition of congruency :
In $\Delta ABC$ and $\Delta QRP , AB =QR ,~ \angle B = \angle R$ and $\angle C= \angle P .$

$\Delta ABC$ and

$\Delta QRP$

$\angle B = \angle R$ [Given]

$\angle C = \angle P$ [Given]

$AB = QR$ [Given]
By Angle - Angle -Side congruency , the triangles

$\Delta ABC$ and

$\Delta QRP$ are congruent to each other

$\Delta ABC \cong \Delta QRP$

1782040_1840838_ans_5aaf720e10e14092afd96a5be0792293.png

In $\Delta$ DEF, DM and EN are two medians. Prove that $3(DF + EF) > 2(DM + EN)$ .

In ADEF, DM and EN are two medians.
To prove :

$3(DF + EF) > 2(DM + EN)$
Proof: In

$\Delta$ DMF
DF + MF > DM
(Sum of any two sides is greater than its third side)

$\Rightarrow DF =\frac{1}{2}EF >DM$ ....(i)
(

$\therefore$ M ia mid-point of EF)
Similarly in

$\Delta$ EFN,
EF +NF>EN

$\Rightarrow EF +\frac{1}{2}DF>EN$ ......(ii)
Adding (i) and (ii),

$(FD +\frac{1}{2}EF + EF +\frac{1}{2}DF) > (DM+ EN)$

$\Rightarrow \frac{3}{2}$ (FD +

$\frac{3}{2}$ EF) > (DM + EN)

$\Rightarrow \frac{3}{2}$ (FD + EF) > (DM) + EN)

$\Rightarrow 3(FD + EF) > 2(DM + EN)$

which of the following pairs of triangle are congruent ? In each case , state the condition of congruency :
In $\Delta ABC$ and $\Delta DEF, AB = DE , BE = EF$ and $\angle B = \angle E.$

$\Delta ABC$ and

$\Delta DEF$

$AB = DE$ [Given]

$\angle B = \angle E$ [Given]

$BC = EF$ [Given]
By side - Angle - side criteion of congruent , the triangles

$\Delta ABC$ and

$\Delta DEF$ are congruent to each other ,

$\Delta ABC \cong \Delta DEF$

1782022_1840832_ans_d0243e3d55f4480d900d469f300bedfc.png

In Fig. which paise of triangles are congruent by SAS congruence criterion (condition)? If congruent, write the congruence of the two triangles in symbolic form.

In triangle

$SUT$ and

$XYZ$ ,

Given,

$XZ = ST$

$YZ = UT = 4cm$

$\angle XZY = \angle STU = 30^{o}$

$\therefore$

$\triangle XZY \cong \triangle SUT$ ...SAS criterion

In Fig. which paise of triangles are congruent by SAS congruence criterion (condition)? If congruent, write the congruence of the two triangles in symbolic form.

In triangle

$DOF$ and

$HOC$ ,

Given,

$HO = DO$

$CO = FO$

$\angle DOF = \angle HOC$ [Vertical opposite angle]

$\therefore$

$\triangle HOC \cong \triangle DOF$ ... SSS criterion

In Fig. which paise of triangles are congruent by SAS congruence criterion (condition)? If congruent, write the congruence of the two triangles in symbolic form.

In triangle

$LMN$ and

$OMN$ ,

Given,

$NM = MN$

$LM = OM$

$\angle LMN = \angle OMN = 40^{o}$

$\therefore$

$\triangle OMN \cong \triangle LMN$ ... SAS criterion

In Fig. which paise of triangles are congruent by $SAS$ congruence criterion (condition)? If congruent, write the congruence of the two triangles in symbolic form.

In triangle

$POR$ and

$TUS$ ,

Given,

$TU = PQ = 3cm$

$US = OR = 5.5cm$

$\angle POR = \angle TUS = 40^{o}$

$\therefore$

$\triangle TUS \cong \triangle POR$ ...SAS criterion

In $\Delta ABC$ and $\Delta DEF \angle B = \angle E= 90^{o}: AC = DF$ and $BC = EF$

$\Delta ABC$ and

$\Delta DEF$

$\angle B = \angle E =90^{o}$
Hyp ,

$AC =$ Hyp

$DF$

$BC = EF$
By Right Angle -Hypotenuse - Side congruent the triangle .

$\Delta ABC$ and

$\Delta DEF$ are congruent to each other .

$\Delta ABC \cong \Delta DEF$

1782030_1840834_ans_50d346f7b6c745a79a945b20bb2469a0.png

In Fig. which paise of triangles are congruent by SAS congruence criterion (condition)? If congruent, write the congruence of the two triangles in symbolic form.

Both triangle

$JKL$ and

$MNO$ are not congruent with any criterion,

In Fig. which paise of triangles are congruent by SAS congruence criterion (condition)? If congruent, write the congruence of the two triangles in symbolic form.

In triangle

$PSQ$ and

$RQS$ ,

Given,

$QS = SQ$

$PS = RQ = 4cm$

$\angle PSQ = \angle RQS = 40^{o}$

$\therefore$

$\triangle PSQ \cong \triangle RQS$ ... SAS criterion

From the following figure: prove that
$AB + AC > BC$

(i)

$\angle ADC +\angle ADB =180^o$ [

$BDC$ is a straight line]

$\angle ADC =90^o$

$90^o +\angle ADB =180^o$

$\angle ADB =90^o .....(i)$
In

$\triangle ADB$ ,

$\angle ADB =90^o$

$\therefore \angle B +\angle BAD =90^o$
Therefore,

$\angle B$ and

$\angle BAD$ ate both acute that less than

$90^o$ .

$\therefore AB > BD.....(ii)$ Side opposite

$90^o$ angle is greater than side opposite acute angle
(ii) In

$\triangle ADC$

$\angle ADB =90^o$

$\therefore \angle C+\angle DAC =90^o$
Therefore

$\angle C$ and

$\angle DAC$ are both acute, that is less than

$90^o$ .

$\therefore AC > CD....(iii)$ Side opposite

$90^o$ angle is greater than side opposite acute angle
Adding (ii) and (ii)

$AB +AC > BD +CD$

$\Rightarrow AB +AC > BC$

$D$ is a point in side $BC$ of triangle $ABC$ . If $AD > AC$ show that $AB > AC$ .

$AD > AC$ (given)

$\Rightarrow \angle C > \angle ADC ...(1)$
Now,

$\angle ADC > \angle B +\angle BAC$ (Exterior Angle Property)

$\Rightarrow \angle ADC > \angle B ....(2)$
From (1) and (2), we have

$\angle C > \angle ADC > \angle B$

$\Rightarrow \angle C > \angle B$

$\Rightarrow AB > AC$
1782058_1840880_ans_dbf2cb3f654d4ca8a8e9e173c7d81206.png

From the following figure: prove that
$AC > CD$

$\triangle ADC$

$\angle ADB =90^o$

$\therefore \angle C+\angle DAC =90^o$
Therefore

$\angle C$ and

$\angle DAC$ are both acute, that is less than

$90^o$ .

$\therefore AC > CD....(iii)$ Side opposite

$90^o$ angle is greater than side opposite acute angle

In the following figure, $ABC$ is an equilateral triangle $P$ is any point in $AC$ ; prove that:
$BP > PA$

$\triangle ABC$

$AB =BC =CA$ [

$ABC$ is an equilateral triangle]

$\therefore \angle A=\angle B =\angle C$

$\therefore \angle A=\angle B=\angle C =\dfrac {180^o}{3}$

$\Rightarrow \angle A=\angle B=\angle C=60^o$
In

$\triangle ABP$

$\angle A =60^o$

$\angle ABP < 60^o$

$\therefore \angle A > \angle ABP$

$\Rightarrow BP > PA$
[Side opposite to greater side is greater]

From the following figure: prove that
$AB > BD$

$\angle ADC +\angle ADB =180^o$ [

$BDC$ is a straight line]

$\angle ADC =90^o$

$90^o +\angle ADB =180^o$

$\angle ADB =90^o .....(i)$
In

$\triangle ADB$ ,

$\angle ADB =90^o$

$\therefore \angle B +\angle BAD =90^o$
Therefore,

$\angle B$ and

$\angle BAD$ ate both acute that less than

$90^o$ .

$\therefore AB > BD.....(ii)$ Side opposite

$90^o$ angle is greater than side opposite acute angle

In the following figure:
$AC=CD; \angle BAD =1100$ and $\angle ACB =740$
Prove that: $BC > CD$

$\angle ACB =74^o ,,,,(i)$

$\angle ACB +\angle ACD=180^o$ [

$BCD$ is a straight line]

$\Rightarrow 74^o +\angle ACD =180^o$

$\Rightarrow \angle ACD =106^o.......(ii)$
In

$\triangle ACD$ ,

$\angle ACD +\angle ADC +\angle CAD=180^o$
Given that

$AC =CD$

$\Rightarrow \angle ADC =\angle CAD$

$\Rightarrow 106^o +\angle CAD +\angle CAD =180^o$

$\Rightarrow 2\angle CAD =74^o$

$\Rightarrow \angle CAD=37^o =\angle ADC ....(iii)$
Now,

$\angle BAD =110^o$

$\angle BAC +\angle CAD =110^o$

$\angle BAC+37^o =110^o$
\angle BAC =73^o ....In

$\triangle ABC$ ,

$\angle B +\angle BAC +\angle ACB =180^o$

$\Rightarrow \angle B+73^o+74^o=180^o$

$\Rightarrow \angle B +174^o =180^o$

$\Rightarrow \angle B=33^o ....(v)$

$\therefore \angle BAC > \angle B$

$\Rightarrow BC > AC$
But,

$AC -CD$

$\Rightarrow BC > CD$

Which of the following pairs of triangle are congruent ? In each case , state the condition of congruency :
In $\Delta ABC$ and $\Delta PQR , AB =PQ ,AC = PR$ and $BC = QR .$

$\Delta ABC$ and

$\Delta PQR$

$AB = PQ$ [Given]

$AC = PR$ [Given]

$BC = QR$ [Given]
By Side - Side -Side criterion of congruency to each other

$\Delta ABC \cong \Delta PQR$

1782044_1840840_ans_15de86ed357b420883ac304be656a5c5.png

In the following figure, $\angle BAC =60^o$ and $\angle ABC=65^o$
Prove that:
$CF > AF$

$\triangle BEC$

$\angle B +\angle BEC +\angle BCE =180^o$

$\angle B=65^o$

$\angle BEC =90^o$ [

$CE$ is perpendicular to

$AB$ ]

$\Rightarrow 65^o +90^o +\angle BCE=180^o$

$\Rightarrow \angle BCE=180^o -155^o$

$\Rightarrow \angle BCE =25^o =\angle DCF....(i)$
In

$\triangle CDF$ ,

$\angle DCF +\angle FDC +\angle CFD =180^o$

$\angle DCF =25^o$

$\angle FDC =90^o$ [

$AD$ i sperpendicular to

$BC$ ]

$\Rightarrow 25^o +90^o +\angle CFD=180^o$

$\Rightarrow \angle CFD =180^o -115^o$

$\Rightarrow \angle CFD =65^o ....(ii)$
Now,

$\angle AFC +\angle CFD=180^o$ [

$AFD$ is a straight line]

$\Rightarrow \angle AFC +65^o =180^o$

$\Rightarrow \angle AFC =115^o ........(iii)$
In

$\triangle ACE$

$\angle ACE +\angle CEA +\angle BAC =180^o$

$\angle BAC =60^o$

$\angle CEA =90^o$ [

$CE$ is perpendicular to

$AB$ ]

$\Rightarrow \angle ACE +90^o +60^o =180^o$

$\Rightarrow \angle ACE =180^o -150^o$

$\Rightarrow \angle ACE =30^o .....(iv)$
In

$\triangle AFC$ ,

$\angle AFC +\angle ACF +\angle FAC =180^o$

$\angle AFC =115^o$

$\angle ACF =30^o$

$\Rightarrow 115^o +30^o +\angle FAC =180^o$

$\Rightarrow \angle FAC =180^o -145^o$

$\Rightarrow \angle FAC =35^o ......(v)$
In

$\triangle AFC$

$\angle FAC =35^o$

$\angle ACF =30^o$

$\therefore \angle FAC > \angle ACF$

$\Rightarrow CF > AF$

In the following figure, $\angle BAC =60^o$ and $\angle ABC=65^o$
Prove that:
$DC > DF$

$\triangle BEC$

$\angle B +\angle BEC +\angle BCE =180^o$

$\angle B=65^o$

$\angle BEC =90^o$ [

$CE$ is perpendicular to

$AB$ ]

$\Rightarrow 65^o +90^o +\angle BCE=180^o$

$\Rightarrow \angle BCE=180^o -155^o$

$\Rightarrow \angle BCE =25^o =\angle DCF....(i)$
In

$\triangle CDF$ ,

$\angle DCF +\angle FDC +\angle CFD =180^o$

$\angle DCF =25^o$

$\angle FDC =90^o$ [

$AD$ i sperpendicular to

$BC$ ]

$\Rightarrow 25^o +90^o +\angle CFD=180^o$

$\Rightarrow \angle CFD =180^o -115^o$

$\Rightarrow \angle CFD =65^o ....(ii)$
Now,

$\angle AFC +\angle CFD=180^o$ [

$AFD$ is a straight line]

$\Rightarrow \angle AFC +65^o =180^o$

$\Rightarrow \angle AFC =115^o ........(iii)$
In

$\triangle ACE$

$\angle ACE +\angle CEA +\angle BAC =180^o$

$\angle BAC =60^o$

$\angle CEA =90^o$ [

$CE$ is perpendicular to

$AB$ ]

$\Rightarrow \angle ACE +90^o +60^o =180^o$

$\Rightarrow \angle ACE =180^o -150^o$

$\Rightarrow \angle ACE =30^o .....(iv)$
In

$\triangle AFC$ ,

$\angle AFC +\angle ACF +\angle FAC =180^o$

$\angle AFC =115^o$

$\angle ACF =30^o$

$\Rightarrow 115^o +30^o +\angle FAC =180^o$

$\Rightarrow \angle FAC =180^o -145^o$

$\Rightarrow \angle FAC =35^o ......(v)$
In

$\triangle AFC$

$\angle FAC =35^o$

$\angle ACF =30^o$

$\therefore \angle FAC > \angle ACF$

$\Rightarrow CF > AF$
In

$\triangle CDF$ ,

$\angle DCF =25^o$

$\angle CFD =65^o$

$\therefore \angle CFD > \angle DCF$

$\Rightarrow DC > DF$

Which of the following pairs of triangle are congruent ? In each case , state the condition of congruency :
In $\Delta ABC$ and $\Delta PQR , BC = QR = \angle A = 90^{o} , \angle C= \angle R = 40 ^{o}$ and $\angle Q = 50^{o}$

$\Delta PQR$

$\angle R = 40^{o} , \angle Q = 50^{o}$

$\angle \angle P +\angle Q +\angle R = 180^{o} \left [ Sum of all the angles in a triangle = 180^{0} \right ]$

$\Rightarrow \angle P + 50^{o} + 40^{o} = 180^{o}$

$\Rightarrow \angle P + 90^{o} = 180^{o}$

$\Rightarrow \angle P = 180^{o} - 90^{o}$

$\Rightarrow \angle P = 90^{o}$
In

$\Delta ABC$ and

$\Delta PQR$

$\angle A = \angle P$

$\angle C = \angle R$

$BC = QR$
By Angle - Angle - Side crition of congruency , the triangle

$\Delta ABC$ and

$\Delta PQR$ are congruent to each other

$\therefore \Delta ABC \cong \Delta PQR$

1782047_1840842_ans_7000735851154e02bf682ccc492c89ff.png

In the following figure, $ABC$ is an equilateral triangle $P$ is any point in $AC$ ; prove that:
$BP > PC$

$\triangle BPC$

$\angle C=60^o$

$\angle CBP < 60^o$

$\therefore \angle C > \angle CBP$

$\Rightarrow BP > PC$
[Side opposite to greater side is greater]

$P$ is any point inside the triangle $ABC$ . Prove that $\angle BPC > \angle BAC$ .

Let

$\angle PBC =x$ and

$\angle PCB=y$
then,

$\angle BPC =180^o - (x+y)....(i)$
Let

$\angle ABP =a$ and

$\angle ACP =b$
then,

$\angle BAC =180^o -(x+a)-(y+b)$

$\Rightarrow \angle BAC =180^o -(x+y)-(a+b)$

$\Rightarrow \angle BAC =\angle BPC -(a+b)$

$\Rightarrow \angle BPC =\angle BAC + (a+b)$

$\Rightarrow \angle BPC > \angle BAC$

1782193_1840912_ans_ecea50528a3247a5930ea10b3e30cfe1.png

The sides $AB$ and $AC$ of a triangle $ABC$ are produced; and the bisects of the external angles at $B$ and $C$ meet at $P$ . Prove that if $AB > AC$ , then $PC > PB$ .

$\triangle ABC$ ,

$AB > AC$

$\Rightarrow \angle ABC < \angle ACB$

$\therefore 180^o - \angle ABC > 180^o -\angle ACB$

$\Rightarrow \dfrac {180^o -\angle ABC}{2} > \dfrac {80^o -\angle ACB}{2}$

$\Rightarrow 90^o -\dfrac 12 \angle ABC > 90^o -\dfrac 12 \angle ACB$

$\Rightarrow \angle CBP > \angle BOP$ [

$BP$ is bisector of

$\angle CBD$ and

$CP$ is bisector to

$\angle BCE$ ]

$\Rightarrow PC > PB$ [side opposite to greater angle is greater]

1782200_1840927_ans_1f1647d9836f4f3ca90121c38a7d0c44.png

In quadrilateral $ABCD$ , side $AB$ is the longest and side $DC$ is the shortest. Prove that:
$\angle C > \angle A$

In the quad

$ABCD$
Since

$AB$ is the longest side and

$DC$ is the shortest side:
(i)

$\angle 1 > \angle 2 [AB > BC]$

$\angle 7> \angle 4 [AD > DC]$

$\therefore \angle 1+\angle 7 > \angle 2+\angle 4$

$\Rightarrow \angle C > \angle A$

1782214_1840938_ans_c640e4ad626b4deb8048821dd9e0ba3e.png

In a triangle ABC, D is the midpoint of BC; AD is produced up to E so that DE = AD prove that:
$\angle DAB = \angle DEC$

Given:

$\Delta ABC$ in which D is the midpoint of BC
AD is produced to E so that DE = AD
We need to prove that

(1)

$\Delta ABD \cong \Delta ECD$

(2) AB = EC

$\Delta ABD$ and

$\Delta ECD$ ,
BD = DC [D is the midpoint of BC]

$\angle ADB = \angle CDE$ [Vertically opposite angles]
AD = DE [Given]
By Side - Angle - Side criterion of congruence we have

$\Delta ABD \cong \Delta ECD$

Also,

$\angle DAB = \angle DEC$ [The corresponding parts of congruent triangles are equal]

1782213_1840937_ans_b4f6f8c3d5c84f06a768bb4e92473d8e.png

In the given figure, $AB=AC.$ Prove that:
$DP=DQ$

Construction :
join

$AD.$
In

$\triangle ABC,$

$AB=AC$ [Given]

$\therefore \angle C=\angle B$ ....(i) [angles opp.to equal sides are equal]

In

$\triangle BPD$ and

$\triangle CQD,$

$\angle BPD=\angle CQD$ [Each

$=90^{o}$ ]

$\angle B=\angle C$ [Proved]

$BD=DC$ [Given]

$\therefore \triangle BPD\cong \triangle CQD$ [

$AAS$ criterian]

$\therefore DP=DQ$ [

$cp ct$ ]

1782218_1840940_ans_acd6a059d10e41cf93ce8d24207516d6.png

In isosceles triangle $ABC$ sides, $AB$ and $AC$ are equal. If point $D$ lies in base $BC$ and point $E$ lies on $BC$ produced ( $BC$ being produced to vertex $C$ ). prove that:
$AC > AD$

We know that the bisector of the angle at the vertex of an triangle bisects the base at right angle.
Using Pythagoras theorem i

$\triangle AFB$

$AB^2 =AF^2+BF^2 ....(i)$
In

$\triangle AFD$ ,

$AD^2 =AF^2 +DF^2 .....(ii)$
We know

$ABC$ is isosceles triangle and

$AB=AC$

$AC^2 =AF^2 +BF^2 .....(iii)$
Subtracting (ii) from (iii)

$AC^2 -AD^2 =AF^2+BF^2-AF^2-DF^2$

$AC^2-AD^2 =BF^2-DF^2$
Let

$2DF =BF$

$AC^2 -AD^2=(2DF)^2 - DF^2$

$AC^2-AD^2=4DF^2-DF^2$

$AC^2=AD^2+3DF^2$

$\Rightarrow AC^2 > AD^2$

$\Rightarrow AC > AD$
Similarly

$AE > AC$ and

$AE > AD$ .
1782235_1840948_ans_112c7d1b1a7643c88e5ebbc3837bae66.png

Prove that the straight line joining the vertex of an isosceles triangle to any point in the base is smaller than either of the equal sides of the triangle.

We know that exterior angle of a triangle is always greater than each of the interior opposite angles.

$\therefore$ In

$\triangle ABD$

$\angle ADC > \angle B ....(i)$
In

$\triangle ABC$ ,

$AB =AC$

$\therefore \angle B =\angle C....(ii)$
From (i) and (ii)

$\angle ADC > \angle C$
(i) In

$\triangle ADC$

$\angle ADC > \angle C$

$\therefore AC > AD .....(iii)$ [side opposite to greater angle is greater]
(ii) In

$\triangle ABC$ ,

$AB =AC$

$\Rightarrow AB > AD$

1782195_1840917_ans_250a17b9cda24f78a7f09a181acfbf4c.png

In a triangle ABC , D is mid - point of BC ; AD is produced upto E so that DE = AD prove that :
$\Delta ABD$ and $\Delta ECD$ are congruent

Given ; A

$\Delta ABC$ in which D is the mid - point of BC
AD is produce to E so that DE = AD
We need to prove that

$\Delta ABD \cong \Delta ECD$
AB = EC
AB \\ EC
In

$\Delta ABD$ and

$\Delta ECD$
BD = DE [D is the midpoint of BC]

$\angle ADB = \angle CDE$ [Vertically opposite angles]
AD = DE [Given]
By Side - Angle - Side criterion of congruence we have

$\Delta ABD \cong \Delta ECD$
1782199_1840924_ans_b3bcc20337724dec93aeb91eca15bab7.png

In a triangle ABC , D is mid - point of BC ; AD is produced upto E so that DE = AD prove that :
AB = EC

Given ; A

$\Delta ABC$ in which D is the mid - point of BC
AD is produced to E so that DE = AD
We need to prove that

$\Delta ABD \cong \Delta ECD$

AB = EC
AB \\ EC
In

$\Delta ABD$ and

$\Delta ECD$
BD = DE [D is the midpoint of BC]

$\angle ADB = \angle CDE$ [Vertically opposite angles]
AD = DE [Given]
By Side - Angle - Side criterion of congruence we have

$\Delta ABD \cong \Delta ECD$

AB = EC [The corresponding parts of congruent triangles are equal]

1782205_1840930_ans_12b5579b05864a0ab1c9c9fc1b581c2d.png

In isosceles triangle $ABC$ sides, $AB$ and $AC$ are equal. If point $D$ lies in base $BC$ and point $E$ lies on $BC$ produced ( $BC$ being produced vertex $C$ ). prove that:
$AE > AC$

We know that the bisector of the angle at the vertex of an triangle bisects the base at right angle.
Using Pythagoras theorem i

$\triangle AFB$

$AB^2 =AF^2+BF^2 ....(i)$
In

$\triangle AFD$ ,

$AD^2 =AF^2 +DF^2 .....(ii)$
We know

$ABC$ is isosceles triangle and

$AB=AC$

$AC^2 =AF^2 +BF^2 .....(iii)$
Subtracting (ii) from (iii)

$AC^2 -AD^2 =AF^2+BF^2-AF^2-DF^2$

$AC^2-AD^2 =BF^2-DF^2$
Let

$2DF =BF$

$AC^2 -AD^2=(2DF)^2 - DF^2$

$AC^2-AD^2=4DF^2-DF^2$

$AC^2=AD^2+3DF^2$

$\Rightarrow AC^2 > AD^2$

$\Rightarrow AC > AD$
Similarly

$AE > AC$ and

$AE > AD$ .
1782236_1840949_ans_d8d2df75bc0143b08c3dc08d89008b11.png

In the given figure $AB / FD , AC/GE$ and $BD = CE$ : prove that :
$CF = EG$

In the given figure AB || FD ,

$\Rightarrow \angle ABC = \angle FDC$
Also AC || GE

$\Rightarrow \angle ACB = \angle GEB$
Consider the two triangles

$\Delta GBE$ and

$\Delta FDC$

$\angle B = \angle D$

$\angle C = \angle E$
Also given that
BD = CE

$\Rightarrow BD + DE = CE + DE$

$\Rightarrow BE = DC$
By Angle - side - Angle criterion of congruence

$\Delta GBE \cong \Delta FDC$

$\therefore \dfrac{GB}{FD} = \dfrac{BE}{DC} = \dfrac{GE}{FC}$

But

$BE = DC \Rightarrow \dfrac{BE}{DC} = \dfrac{BE}{BE} = 1$

$\therefore \dfrac{GB}{FD} = \dfrac{BE}{DC} = 1$

$\Rightarrow GB = FD$

$\therefore \dfrac{GE}{FC} = \dfrac{BE}{DC} = 1$

$\Rightarrow GE = FC$

In isosceles triangle $ABC$ sides, $AB$ and $AC$ are equal. If point $D$ lies in base $BC$ and point $E$ lies on $BC$ produced ( $BC$ being produced vertex $C$ ). prove that:
$AE > AD$

We know that the bisector of the angle at the vertex of an triangle bisects the base at right angle.
Using Pythagoras theorem i

$\triangle AFB$

$AB^2 =AF^2+BF^2 ....(i)$
In

$\triangle AFD$ ,

$AD^2 =AF^2 +DF^2 .....(ii)$
We know

$ABC$ is isosceles triangle and

$AB=AC$

$AC^2 =AF^2 +BF^2 .....(iii)$
Subtracting (ii) from (iii)

$AC^2 -AD^2 =AF^2+BF^2-AF^2-DF^2$

$AC^2-AD^2 =BF^2-DF^2$
Let

$2DF =BF$

$AC^2 -AD^2=(2DF)^2 - DF^2$

$AC^2-AD^2=4DF^2-DF^2$

$AC^2=AD^2+3DF^2$

$\Rightarrow AC^2 > AD^2$

$\Rightarrow AC > AD$
Similarly

$AE > AC$ and

$AE > AD$ .
1782239_1840950_ans_55b242f0580b4180a9609c58f316ebd3.png

Given: $ED=EC$
Prove : $AB + AD > BC$

The sum of nay two sides of the triangle is always greater than the third side of the triangle.
In

$\triangle CEB$

$CE +EB > BC$

$\Rightarrow DE +EB > BC \quad [CE=DE]$

$\Rightarrow DB > BC ....(i)$
In

$\triangle ADB$ ,

$AD+AB > BD$

$\Rightarrow AD+AB > BD < BC$

$\Rightarrow AD +AB > BC$ .

In triangle $ABC, AB > AC$ and $D$ is a point in side $BC$ . Show that: $AB > AD$ .

Given that,

$AB > AC$

$\Rightarrow \angle C > \angle B.....(i)$
Also In

$\triangle ADC$

$\angle ADB=\angle DAC +\angle C$

$\Rightarrow \angle ADB > \angle C$

$\Rightarrow \angle ADB > \angle C > \angle B$

$\Rightarrow \angle ADB > \angle B$

$\Rightarrow AB > AD$

1782247_1840957_ans_c811407a49af4be4aa80121a20d8972f.png

In the following figure AB = AC and AD is perpendicular to BC . BE bisects angle B and EF is perpendicular to AB
Prove that
ED = EF

$\Delta EFB$ and

$\Delta EDB$

$\angle EFB = \angle EDB$ ( both are

$90 ^{o}$ )

$EB = EB$ [ common ]

$\angle FBD = \angle DBE$ (given )

$\Delta EFB \cong \Delta EDB$ By angle-side-angle congruency

Also

$ED = EF$ (By C.P.C.T)

In the given figure, $AB=AC.$ Prove that:
$AP=AQ$

Draw a line joining

$A$ with the midpoint of

$BC.$

$\triangle ABC,$

$AB=AC$ [Given]

$\therefore \angle C=\angle B$ ...(I) [angles opp.to equal sides are equal]

$\triangle BPD$ and

$\triangle CQD$ , we have:

$BD=DC$ [Given]

$\angle C=\angle B$ [From (I)]

$\angle BPD=\angle CQD$ [Each

$=90^0$ ]

So,

$\triangle BPD\cong \triangle CQD$

Also,

$BP=CQ$ [cpct]

Now,

$AB=AC$ [Given]

$\Rightarrow AB-BP=AC-CQ$

$\Rightarrow AP=AQ$

1784377_1841066_ans_14a9e05c79504aecb79ba455ebc706e7.jpg

Use the information in the given figure to prove ;
AB = FE

$\Delta ABC$ and

$\Delta EFD$

$AB || EF$

$\Rightarrow \angle ABC = \angle EFD$ (alternate angles )

$AC = ED$ (given )

$\angle ACB = \angle EFD$ (given)

$\Delta ACB \cong \Delta EFD$ ( AAS congruence criterion )

$\Rightarrow AB = FE$ (cpct)

A triangles ABC has $\angle B = \angle C$
Prove that
The perpendicular from the mid - point of BC to AB and AC are equal'

Given ; A

$\Delta ABC in which \angle B = \angle C .$
DL is the perpendicular from D to AB
DM is the perpendicular from A to AC
We need to prove that.
DL = DM
Proof ;
In

$\Delta DLB$ and

$\Delta DMC$

$\angle DLB = \angle DMC = 90^{o}$

$[DL \perp AB$ and

$DM \perp AC]$

$\angle B = \angle C$ [Given]

$BD = DC$ [D is the mid point of BC]
By Angle - Angle - Side criterion of congruence

$\Delta DLB \cong \Delta DMC$
The corresponding parts of the congruent triangles are congruent

$DL = DM$ [c.p.c.t]
1782251_1840961_ans_c31fc5b857ca41de8cf91422cf52300d.png

A triangles ABC has $\angle B = \angle C$
Prove that
The perpendicular from B and C to the opposite sides are equal .

Given ; A

$\Delta ABC$ in which

$\angle B = \angle C$
BP is the perpendicular from D to AC
CQ is the perpendicular from C to AB
We need to prove that
BP = CQ
Proof
In

$\Delta BPC$ and

$\Delta CQB$

$\angle B = \angle C$ [Given]

$\angle BPC = \angle CQB = 90^{o} [BP \perp AC and CQ \perp AB ]$
BC = BC [common]
By Angle - Angle - Side criterion of congruence ,

$\Delta BPC \cong \Delta CQB$
The corresponding parts of the congruent ;triangle are congruent
BP = CQ [c . p . c. t ]
1782258_1840970_ans_3e5de1fc7d9e4523afa24092cfda86f2.png

1782258_1840970_ans_3e5de1fc7d9e4523afa24092cfda86f2.png

Use the information in the given figure to prove ;
BD = CF

$\Delta ABC$ and

$\Delta EFD$

$AB || EF \Rightarrow \angle ABC = \angle EFD$ (alternate angles )

AC = ED (given )

$\angle ACB = \angle EFD$ (given)

$\Delta ACB \cong \Delta EFD$ ( AAS congruence criterion )

Also,

$BC = DF$ (cpct)

In the figure given below triangles $ABC$ is right-angled at $B .\ ABPQ$ and $ACRS$ are squares. Prove that :
$\Delta ACQ$ and $\Delta ASB$ are congruent

Given ; A

$\Delta ABC$ is right-angled at

$B .$

$ABPQ$ and

$PQRS$ are square
We need to prove that

$\Delta ACQ \cong \Delta ASB$

$\angle QAB = 90^{0}$

$\angle SAC = 90^{o}$

$\angle QAB = \angle SAC$
Adding

$\angle BAC$ to both sides of (3) , we have

$\angle QAB + \angle BAC = \angle SAC +\angle BAC$

$\Rightarrow \angle QAC = \angle SAB$ ...(1)

Now, In

$\Delta ACQ$ and

$\Delta ASB$ ,

$QA=QB$ ...[Sides of a square ABPQ]

$\angle QAC=\angle SAB$ ...[From (1)]

$AC=AS$ ....[Sides of a squre ACRS]
So, By angle-angle-side criterion of congruence,

$\Delta ACQ \cong \Delta ASB$

In the figure given below triangles ABC is right - angled at B . ABPQ and ACRS are square . Prove that :
CQ = BS

Given ; A

$\Delta ABC$ is right-angled at

$B .$

$ABPQ$ and

$PQRS$ are square
We need to prove that

$\Delta ACQ \cong \Delta ASB$

$\angle QAB = 90^{0}$

$\angle SAC = 90^{o}$

$\angle QAB = \angle SAC$
Adding

$\angle BAC$ to both sides of (3) , we have

$\angle QAB + \angle BAC = \angle SAC +\angle BAC$

$\Rightarrow \angle QAC = \angle SAB$ ...(1)

Now, In

$\Delta ACQ$ and

$\Delta ASB$ ,

$QA=QB$ ...[Sides of a square ABPQ]

$\angle QAC=\angle SAB$ ...[From (1)]

$AC=AS$ ....[Sides of a squre ACRS]
So, By angle-angle-side criterion of congruence,

$\Delta ACQ \cong \Delta ASB$

The corresponding parts of the congruent triangles are congruent

$CQ = BS$ [c.p.c. t]

Use the information in the given figure to prove ;
BC = DF

$\Delta ABC$ and

$\Delta EFD$

$AB || EF \Rightarrow \angle ABC = \angle EFD$ (alternate angles )

$AC = ED$ (given )

$\angle ABC = \angle EFD$ ( alternate angles )

$\Delta ACB \cong \Delta EFD$ ( AAS congruence criterion )

And

$BC = DF$ (cpct)

Adding

$DC$ we get:

$\Rightarrow BD + DC = CF + DC$

$\Rightarrow BC = DF$

In triangle $ABC$ ; $AB=AC.\ P,Q$ and $R$ are mid-points of sides $AB,AC$ and $BC$ respectively. Prove that:
$PR=QR$

$\triangle ABC,$

$AB=AC$ [Given]

$\Rightarrow \dfrac{1}{2}AB=\dfrac{1}{2}AC$

$\Rightarrow AP=AQ$ ......(i) [Since

$P$ and

$Q$ are mid-points]
In

$\triangle BCA,$

$PR=\dfrac{1}{2}AC$ [

$PR$ is line joining the mid-points of

$AB$ and

$BC$ ]

$\Rightarrow PR=AQ$ ......(ii)
In

$\triangle CAB,$

$QR=\dfrac{1}{2}AB$ [

$QR$ is line joining the mid-points of

$AC$ and

$BC$ ]

$\Rightarrow QR=AP$ ......(iii)
From (i)(ii) and (iii)

$PR=QR$

1785637_1841188_ans_87a122196b2146dd8cc6f8a7e6436760.png

In the given figure, $AB=AC$ and $\angle DBC=\angle ECB=90^{o}.$
$BD=CE$

$\triangle ABC,$

$AB=AC$ [Given]

$\therefore \angle ACB=\angle ABC$ [angles opp. to equal sides are equal]

$\Rightarrow \angle ABC=\angle ACB$ .....(i)

$\angle DBC=\angle ECB=90^{o}$ [Given]

$\Rightarrow \angle DBC=\angle ECB$ .......(ii)
Subtracting (i) from (ii)

$\angle DCB-\angle ABC=\angle ECB-\angle ACB$

$\Rightarrow \angle DBA=\angle ECA$ ....(iii)
In

$\triangle DBA$ and

$\triangle ECA,$

$\angle DBA=\angle ECA$ [From (iii)]

$\angle DAB=\angle EAC$ [Vertically opposite angles]

$AB=AC$ [Given]

$\triangle DBA\cong \triangle ECA$ [ASA]

$\Rightarrow BD=CE$ [cpct]

In triangle $ABC$ ; $AB=AC.\ P,Q$ and $R$ are mid-points of sides $AB,AC$ and $BC$ respectively. Prove that:
$BQ=CP$

$AB=AC$ [Given]

$\Rightarrow \angle B=\angle C$
Also,

$\dfrac{1}{2}AB=\dfrac{1}{2}AC$

$\Rightarrow BP=CQ$ [

$P$ and

$Q$ are mid- points of

$AB$

$AC$ ]
In

$\triangle BPC$ and

$\triangle CQB,$

$BP=CQq$

$\angle B=\angle C$

$BC=BC$
Hence,

$\triangle BPC\cong\triangle CQB$ [SAS]

$BQ=CP$ [C.P.C.T]

1785640_1841190_ans_d0304d025a3447d4b6a03deaade40d44.png

In the given figure, $AB=AC$ and $\angle DBC=\angle ECB=90^{o}.$
$AD=AE$

$\triangle ABC,$

$AB=AC$ [Given]

$\therefore \angle ACB=\angle ABC$ [angles opp. to equal sides are equal]

$\Rightarrow \angle ABC=\angle ACB$ .....(i)

$\angle DBC=\angle ECB=90^{o}$ [Given]

$\Rightarrow \angle DBC=\angle ECB$ .......(ii)
Subtracting (i) from (ii)

$\angle DCB-\angle ABC=\angle ECB-\angle ACB$

$\Rightarrow \angle DBA=\angle ECA$ ....(iii)
In

$\triangle DBA$ and

$\triangle ECA,$

$\angle DBA=\angle ECA$ [From (iii)]

$\angle DAB=\angle EAC$ [Vertically opposite angles]

$AB=AC$ [Given]

$\triangle DBA\cong \triangle ECA$ [ASA]
Also,

$AD=AE$ [cpct]

From the following figure, Prove that:
$AD=CE$

$\triangle ACB,$

$AC=AC$ [Given]

$\therefore \angle ABC=\angle ACB$ ......(i) [angles opp. to equal sides are equal]

$\angle ACD=\angle ACB=180^{o}$ ......(ii) [

$DCB$ is a straight line ]

$\angle ABC=\angle CBE=180^{o}$ ......(iii) [

$ABE$ is a straight line ]
Equating (ii) and (iii)

$\angle ACD+\angle ACB=\angle ABC+\angle CBE$

$\angle ACD+\angle ACB=\angle ACB+\angle CBE$ [From (i)]

$\Rightarrow \angle ACD=\angle CBE$

$\triangle ACD$ and

$\triangle CBE,$

$DC=CB$ [Given]

$AC=BE$ [Given]

$\angle ACD=\angle CBE$ [Proved Earlier]

$\therefore \triangle ACD\cong \triangle CBE$ [SAS criterian]

$\Rightarrow AD=CE$ [cpct]

From the following figure, Prove that:
$\angle ACD=\angle CBE$

$\triangle ACB,$

$AC=AC$ [Given]

$\therefore \angle ABC=\angle ACB$ ......(i) [angles opp. to equal sides are equal]

$\angle ACD=\angle ACB=180^{o}$ ......(ii) [

$DCB$ is a straight line ]

$\angle ABC=\angle CBE=180^{o}$ ......(iii) [

$ABE$ is a straight line ]
Equating (ii) and (iii)

$\angle ACD+\angle ACB=\angle ABC+\angle CBE$

$\angle ACD+\angle ACB=\angle ACB+\angle CBE$ [From (i)]

$\Rightarrow \angle ACD=\angle CBE$

In the following figure AB = EF , BC = DE and $\angle B = \angle E = 90^{o}$
Prove that AD = FC

Given that,

$BC = DE$

$\Rightarrow BC + CD = DE + CD$ (Adding

$CD$ on both sides)

$\Rightarrow BD = CE$ ...(1)
Now, In

$\Delta ABD$ and

$\Delta EFC$

$AB = EF$ ...[Given]

$\angle ABD = \angle FEC$ ...[Both

$=90^0$ , perpendiculars]

$BD = CE$ ...[From (1)]

$\Rightarrow ABD \cong \Delta \Delta EFC$ ...[SAS congruence criterion]

$\Rightarrow AD = FC$ ...[cpct]
1785849_1841389_ans_572d00f95a28495f8575d2a70edf2689.png

In the following figure , OA = OC and AB = BC ,
prove that :
$\angle AOB = 90^{o}$

In the figure OA = OC , AB = BC

We need to proved that

$\angle AOB = 90^{o}$

$\Delta ABO$ and

$\Delta CBO$

$AB = BC$ [given ]

$OB = OB$ [common]

$OA=OC$ [given]

By Side - Side - Side criterion of congruence , we have

$\Delta ABO \cong \Delta CBO$
The corresponding parts of the congruent triangles are congruent

$\Rightarrow \angle AOB = \angle COB$ [C.P.C,T] ...(1)

Also,

$\angle AOB + \angle COB=180^0$ [Supplementary angles]

Using (1), we can write:

$\angle AOB + \angle AOB=180^0$

$\Rightarrow 2\angle AOB =180^0$

$\Rightarrow \angle AOB =90^0$

The following figure shows a triangle ABC in which AB = AC . M is a point on AB and N is a point AC such that BM = CN
Prove that
AM = AN

$\Delta ABC ,\ AB = AC ,\ M$ and

$N$ are point on

$AB$ and

$AC$ such that

$BM = CN$

$BN$ and

$CM$ are joined.

In

$\Delta AMC$ and

$\Delta ANB$

$AB = AC$ [given] ...(1)

$BM = CN$ [given] ...(2)
Subracting (2) from (1) we have

$AB - BM - AC - CN$

$AM = AN$

In the adjoining figure If $\angle C = \angle F$ , then AB = and BC _

$AB = DE , BC = EF$

In the following figure , OA = OC and AB = BC ,
prove that :
AD = CD

In the figure

$OA = OC ,\ AB = BC$

We need to prove that

$AC=CD$

$\Delta ABO$ and

$\Delta CBO$

$AB = BC$ [given ]

$OB = OB$ [common]

$OA=OC$ [given]

By Side - Side - Side criterion of congruence , we have

$\Delta ABO \cong \Delta CBO$
The corresponding parts of the congruent triangles are congruent

$AD=CD$ [C.P.C,T]

In triangle $ABC,D$ is a points on $BC$ such that $Ab=AD=BD=DC.$ Show that:
$\angle ADC:\angle C=4:1$

Since

$AB=AD=BD$

$\therefore \triangle ABD$ is an equilateral triangle.

$\therefore \angle ADB=60^{o}$

$\Rightarrow \angle ADC=180^{o}-\angle ADB$

$=180^{o}-60^{o}$

$=120^{o}$
Again in

$\triangle ADC,$

$AD=DC$

$\therefore \angle 1=\angle 2$
But ,

$\angle 1+\angle 2+\angle ADC=180^{o}$

$\Rightarrow 2\angle 1+120^{o}=180^{o}$

$\Rightarrow 2\angle 1=60^{o}$

$\Rightarrow \angle 1=30^{o}$

$\Rightarrow \angle C=30^{o}$

$\therefore \angle ADC:\angle C=120^{o}:30^{o}$

$\Rightarrow \angle ADC:\angle C=4:1$
1785801_1841280_ans_37fa61f304464b9c95ab937e1f736ff0.png

M is a point on side BC of a triangle ABC such that AM is the bisector of $\angle BAC.$ Is it true to say that perimeter of the triangle is greater than 2 AM? If True enter 1, else if False enter 0.

Given,

$\triangle ABC$ ,

$AM$ is bisector of

$\angle BAC$

Now, In

$\triangle ABM$ ,

$AB + BM > AM$ (Sum of two sides of triangle is greater than the third side)

Now, In

$\triangle ACM$ ,

$AC + CM > AM$ (Sum of two sides of triangle is greater than the third side)

Thus, adding both,

$AB + BM + AC+ CM > 2 AM$

$AB + AC + BC > 2 AM$

Show that no triangle has two sides each shorter than its corresponding altitude (from the opposite vertex).

Assume that

$AD>BC$ and

$BE>AC$

then

$AD>BC\ge BE$ (Because hypotenuse of a right angled triangle is greater)

According to our assumption

$BE>AC$

So,

$AD>BC\ge BE>AC$

But,

$AC\ge AD$

So, we got a contradiction.

Our assumption is not possible.

Let a, b, c be the three sides of a triangle. Suppose that $\dfrac{a}{b}=\dfrac{b}{c}=q.$
Prove that $\dfrac{\sqrt{5}-1}{2}<q<\dfrac{\sqrt{5}+1}{2}$

Sum of any two sides is greater than third side

$a + b > c$ ......(2)
We have a - b < c < a + b

$\frac{a}{b}-1<\frac{c}{b}<\frac{a}{b}+1$

$q-1<\frac{1}{q}<q+1$

$q^2 - q < 1 < q^2 + q (since\ q\ > 0)$

$q^2 - q -1< 0, q^2 + q + 1 > 0$

$\frac{1-\sqrt{5}}{2}<q<\frac{1+\sqrt{5}}{2}$ ;

$q<\frac{-1-\sqrt{5}}{2}$ or

$q>\frac{-1+\sqrt{5}}{2}$

$\Rightarrow \frac{\sqrt{5}-1}{2}<q<\frac{\sqrt{5}+1}{2}$

The sides of a triangle have lengths $11, ~15$ and $k,$ where $k$ is an integer. For how many values of $k$ is the triangle obtuse?

Sum of two sides of triangle is more than third side

$\Rightarrow 11 + 15 > k$

$\Rightarrow 26>k \Rightarrow k < 26$
difference of any two sides of triangle is less than third side

$\Rightarrow 15-11<k \Rightarrow 4<k \Rightarrow k>4$

$\Rightarrow 4<k<26$

$\Rightarrow k = 5, 6, 7, ....., 25$
Given triangle is obtuse,

$\Rightarrow$ either

$11^2 + 15^2 < k^2$ or

$11^2 + k^2 < 15^2$

$11^2 + 15^2 < k^2 \Rightarrow k = 19, 20, 21, ...., 25$

$11^2 + k^2 < 15^2 \Rightarrow k = 5, 6, 7, .... 10$
Number of values possible for

$k$ are

$7+6=13$

Show that in any triangle with sides a, b and c, we have $(a + b + c)^2 < 4 (ab + bc + ca).$

Without loss of generality, let us assume that a > b > c > O. Since the sum of any two sides of a triangle is always greater than the third side, therefore the difference between any two sides can never exceed the third and hence

$0 \leq b-c < a, 0 \leq c-a < b, 0 \leq a-b < c$

Squaring both sides of the above inequalities and adding the corresponding sides,

We get

$(b - c)^2 + (a - c)^2 + (a - b)^2 < a^2 + b^2 + c^2$

$\Leftrightarrow a^2 + b^2 + c^2 < 2(bc + ca + ab)$

$\Leftrightarrow (a + b + c)^2 < 4(bc + ca + ab)$

State whether the given statement is true/false:
In the following diagram, $AB = AC$ then which of the following is correct?
$AE>AF$
$AF>AE$

Since, $AB =AC$

Let $\angle B=\angle C=x$

Now, $\angle AEF= \angle B+ \angle D=x+ \angle D$

And $\angle AFE= \angle CFD= \angle C \angle D=x -\angle D$

$\angle AEF> \angle AFE \Rightarrow AF>AE$

In the given figure, $\displaystyle BC=CE$ and $\displaystyle \angle 1=\angle 2$ Then,
: $\displaystyle \Delta GCB \equiv \Delta DCE$
If the above statement is true then mention answer as 1, else mention 0 if false

$\angle CBG = 180^\circ - \angle 1$

$\angle CED = 180^\circ - \angle 2$
Since,

$\angle 1 = \angle 2$
Thus,

$\angle CBG = \angle CED$
Now, In

$\triangle GCB$ and

$\triangle DCE$ ,

$\angle CBG = \angle CED$ (Proved above)

$BC = CE$ (Given)

$\angle GCB = \angle DCE$ (Vertically opposite angles)
Thus,

$\triangle GCB \cong \triangle DCE$ ...... (ASA rule)

$AD$ is perpendicular to $BC$ .

$\triangle ABD$ and

$\triangle ACD$ ,

$AD = AD$ [Common]

$\angle BAD = \angle CAD$ [

$AD$ is the bisector of angle

$A$ ]

$AB = AC$ [Given]

So, by

$SAS$ rule of congruence,

$\triangle ABD \cong \triangle ACD$

As we know that, corresponding angles of congruent triangles are equal so,

$\angle ADC = \angle ADB = x$

Now,

$\angle ADC + \angle ADB = 180^{\circ}$ [Linear Pair]

$\Rightarrow 2x = 180^{\circ}$

$\Rightarrow x = 90^{\circ}$

$\Rightarrow \angle ADC = \angle ADB = 90^{\circ}$

Thus,

$AD \perp BC$

AB + BC + CD > DA
If the above statement is true, then enter 1 or else enter 0.

In $\triangle ABC$ , we have sum of two sides > third side
$=> AB + BC > AC$ -- (1)
Also, in $\triangle ADC$ , we have sum of two sides > third side
$=> DC + AC > AD$ ---- (2)
Adding equations $1, 2$ we get
$=> AB + BC + CD + AC > AC + AD$
$=> AB + BC + CD > AD$

The given figure shows $\displaystyle PQ=PR$ and $\displaystyle \angle Q=\angle R$ Prove that $\displaystyle \Delta PQS \equiv \Delta PRT$

$\triangle PQS$ and

$\triangle PRT$ ,

$PQ = PR$ (Given)

$\angle Q = \angle R$ (Given)

$\angle QPS = \angle RPT$ (Common angle)
Hence,

$\triangle PQS \cong \triangle PRT$ (SAS rule)

The given figure shows a triangle $ABC$ in which $AB = AC$ . $M$ is a point on $AB$ and $N$ is a point on $AC$ such that $BM = CN$ . Hence, $BN=CM$ .
If the above statement is true, then mention answer as 1, else mention 0 if false.

$AB = AC$ (Given)

$BM = NC$ (Given)

Subtracting both the equations:

$AB - BM = AC - NC$

$\implies$

$AM = AN$ .

Now, in

$\triangle$ s,

$ABN$ and

$ACM$ ,

$AB = AC$ (given)

$AM = AN$ (Proved above)

$\angle A = \angle A$ (Common)
Thus,

$\triangle ABN \cong \triangle ACM$ (by

$SAS$ congruency)
Hence,

$BN = CM$ (By

$CPCT$ ).

Therefore, the statement is true and

$1$ is the answer

In triangle ABC, AB > AC and D is a point in side BC.
then, AB > AD.
If the above statement is true then mention answer as 1, else mention 0 if false

Here AB > AC and D is a point in side BC.

here AE is altitude,

So, As per Pythagorean Theorem,

$AD^2=AE^2+ED^2$ ...........eq1

$AB2=AE^2+EB^2$

$AB^2=AE^2+(ED+DB)^2$ ........Eq2

So, As per eq1 & eq2 , it is clearly seen that

$AB>AD$

and any line starting from A inside the triangle ABC, on the base BC will be smaller then longest side.

Answer is 1

$AB$ and $PQ$ bisect each other.
If the above statement is true then mention answer as $1$ , else mention $0$ if false.

Given:

$AP = BQ$ and

$AP \parallel BQ$ .

$\triangle AOP$ and

$\triangle BOQ$ ,

$\angle AOP = \angle BOQ$ (vertically opposite angles)

$AP = BQ$ (Given)

$\angle APO = \angle BQO$ (Alternate angles).
Thus,

$\triangle AOP \cong \triangle BOQ$ (By

$ASA$ rule).

Thus,

$AO = OB$ (By

$CPCT$ )

$OP = OQ$ (By

$CPCT$ )
i.e.

$AB$ bisects

$PQ$ .

Hence, the given statement is true and

$1$ is the answer.

In the given triangle $ABC$ , $D, E$ and $F$ are the mid-points of sides $BC,CA$ and $AB$ respectively. Prove that

$\displaystyle \dfrac{AB-BC}{2}< AE< \dfrac{AB+BC}{2}.$

We know that difference of two sides is less than the third side

$\displaystyle \therefore AB - BC < AC$

Since

$E$ is the mid-point of

$AC$ ,

$AE = \dfrac {AC}{2}$

$\displaystyle AB - BC < 2AE$

$\displaystyle \frac{AB-BC}{2} < AE.....(1)$

We know that the sum of the two sides is greater than the third side

$\displaystyle AB + BC > AC. AC = 2AE. AB + BC > 2AE$

$\displaystyle AE < \frac{AB+BC}{2}.....(2)$

From (1) ad (2),

$\displaystyle \frac{AB-BC}{2}< AE < \frac{AB+BC}{2}.$

$[hence \, proved]$

Show that in an isosceles triangle angles opposite to equal sides are equal

Let

$\displaystyle \bigtriangleup ABC$ be an isosceles triangle such that AB =AC Then we have to prove that

$\displaystyle \angle B=\angle C$ Draw the bisector AD of

$\displaystyle \angle A$ meeting BC in D
Now in triangles ABD and ACD We have

$AB=AC$ (Given)

$\displaystyle \angle BAD = \angle CAD$ (because AD is bisector of

$\displaystyle \angle A$

$AD = AD$ (Common side)
Therefore by SAS congruence condition we have

$\displaystyle \bigtriangleup ABC\cong \bigtriangleup ACD$

$\displaystyle \Rightarrow \angle B=\angle C$
(Corresponding parts of congruent triangles are equal )
1027747_404667_ans_07eacf0f84ac42b4bab83bffab5e42d1.bmp

1027747_404667_ans_07eacf0f84ac42b4bab83bffab5e42d1.bmp

Using triangle inequality theorem check whether the given side lengths $a = 3, b = 5$ and $c = 1$ will form a triangle or not.

According to the triangle inequality theorem the sum of any two side of the triangle must be greater than the measure of the third side.

$A+B>C$

$B+C>A$

$A+C>B$

Here

$A=3, B=5$ and

$C=1$

$\Rightarrow 3+5>1$

$\Rightarrow 5+1>3$

$\Rightarrow 3+1<5$

$\therefore$ third condition is not satisfied according to the theorem, then the given side length will not form a triangle.

Using triangle inequality theorem check whether the given side lengths $a = 4, b = 5$ and $c = 8$ will form a triangle or not.

According to the triangle inequality theorem the sum of any two side of the triangle must be garter than the measure of the third side.

$A+B>C$

$B+C>A$

$A+C>B$

Here A=4, B=5 and C=8

$\Rightarrow 4+5>8$

$\Rightarrow 5+8>4$

$\Rightarrow 4+8>5$

$\therefore$ All the conditions are satisfied according to the theorm then the given side length will form a triangle.

In Fig, $AC = AE, AB = AD$ and $\angle BAD = \angle EAC$ . Show that $BC = DE$

It is given that

$\angle BAD = \angle EAC$

$\angle BAD + \angle DAC = \angle EAC + \angle DAC$ [add

$\angle DAC$ on both sides]

$\therefore \angle BAC = \angle DAE$

$\triangle BAC$ and

$\triangle DAE$

$AB = AD$ (Given)

$\angle BAC = \angle DAE$ (Proved above)

$AC = AE$ (Given)

$\therefore \triangle BAC \cong \triangle DAE$ (By SAS congruence rule)

$\therefore BC = DE$ (By CPCT)

In right angled triangle $ABC$ , right angled at $C, M$ is the mid-point of hypotenuse $AB$ . $C$ is joined to $M$ and produced to a point $D$ such that $DM = CM$ . Point $D$ is joined to point $B$ . Show that:
(i) $\triangle AMC\cong \triangle BMD$
(ii) $\angle DBC$ is a right angle.
(iii) $\triangle DBC\cong \triangle ACB$
(iv) $CM = \dfrac {1}{2} AB$

(i) In

$\triangle AMC$ and

$\triangle BMD$ , we have

$AM = BM$

$\mid$ Since

$M$ is the mid- point of

$AB$

$\angle AMC=\angle BMD$

$\mid$ Vertically opp. angles

and

$CM = MD$

$\mid$ Given

$\therefore$ By SAS criterion of congruence, we have

$\triangle AMC \cong \triangle BMD$

(ii) Now,

$\triangle AMC \cong \triangle BMD$

$\Rightarrow BD=CA$ and

$\angle BDM=\angle ACM$ ... (1)

$\mid$ Since corresponding parts of congruent triangles are equal

Thus, transversal

$CD$ cuts

$CA$ and

$BD$ at

$C$ and

$D$ respectively such that the alternate angles

$\angle BDM , \angle ACM$ are equal.

$\therefore , BD \parallel CA.$

$\Rightarrow \angle CBD+\angle BCA = 180^{\circ}$

$\mid$ Since sum of consecutive interior angles are supplementary

$\Rightarrow \angle CBD+ 90^{\circ} = 180^{\circ}$

$\mid$ Since

$\angle BCA = 90^{\circ}$

$\Rightarrow \angle CBD = 180^{\circ} - 90^{\circ}$

$\Rightarrow \angle DBC= 90^{\circ}$

(iii) Now, in

$\triangle DBC$ and

$\triangle ACB$ , we have

$BD = CA$

$\mid$ From (1)

$\angle DBC=\angle ACB$

$\mid$ Since Each

$=90^o$

$BC = BC$

$\mid$ Common

$\therefore$ by SAS criterion of congruence, we have

$\triangle DBC \cong \triangle ACB$

(iv)

$CD = AB$ [Since corresponding parts of congruent triangles are equal]

$M$ is the midpoint of

$CD.$

$\Rightarrow CM=\dfrac{1}{2}CD$

$\Rightarrow CM=\dfrac{1}{2}AB$

$ABCD$ is a quadrilateral in which $AD = BC$ and $\angle DAB = \angle CBA$ . Prove that
(i) $\triangle ABD \cong\triangle BAC$
(ii) $BD = AC$
(iii) $\angle ABD = \angle BAC$

$\triangle ABD$ and

$\triangle BAC$ ,

$AD = BC$ (Given)

$\angle DAB = \angle CBA$ (Given)

$AB = BA$ (Common)

$\therefore \triangle ABD \cong \triangle BAC$ (By SAS congruence rule)

$\therefore BD = AC$ (By CPCT)

And,

$\angle ABD = \angle BAC$ (By CPCT)

In $\triangle ABC, AD$ is the perpendicular bisector of $BC$ . Show that $\triangle ABC$ is an isosceles triangle in which $AB = AC$ .

$\triangle ABD$ and

$\triangle ACD$ , we have

$DB = DC$

$\mid$ Given

$\angle ADB=\angle ADC$

$\mid$ since

$AD \bot BC$

$AD = AD$

$\mid$ Common

$\therefore$ by SAS criterion of congruence, we have.

$\triangle ABD\cong \triangle ACD$

$\Rightarrow AB = AC$

$\mid$ Since corresponding parts of congruent triangles are equal

Hence,

$\triangle$

$ABC$ is isosceles.

In an isosceles triangle $ABC$ , with $AB = AC$ , the bisectors of $\angle B$ and $\angle C$ intersect each other at $O$ . Join $A$ to $O$ . Show that :
(i) $OB = OC$ (ii) $AO$ bisects $\angle A$

(i) In

$\triangle ABC$ , we have

$AB = AC$

$\Rightarrow \angle C=\angle B$

$\mid$ Since angles opposite to equal sides are equal

$\Rightarrow \dfrac{1}{2}\angle B=\dfrac{1}{2}\angle C$

$\Rightarrow \angle OBC=\angle OCB$

$\Rightarrow \angle ABO=\angle ACO$

$\dots(1)$

$\Rightarrow OB = OC$

$\mid$ Since sides opp. to equal ∠s are equal

$\dots(2)$

(ii) Now, in

$\triangle ABO$ and

$\triangle ACO$ , we have

$AB = AC$

$\mid$ Given

$\angle ABO=\angle ACO$

$\mid$ From (1)

$OB = OC$

$\mid$ From (2)

$\therefore$ By SAS criterion of congruence, we have

$\triangle ABO \cong \triangle ACO$

$\Rightarrow \angle BAO=\angle CAO$

$\mid$ Since corresponding parts of congruent triangles are equal

$\Rightarrow$

$AO$ bisects

$\angle A$

$l$ and $m$ are two parallel lines intersected by another pair of parallel lines $p$ and $q$ . Show that $\triangle ABC \cong \triangle CDA$

$\triangle ABC$ and

$\triangle CDA$

$\angle BAC = \angle DCA$ (Alternate interior angles, as

$p \parallel q$ )

$AC = CA$ (Common)

$\angle BCA = \angle DAC$ (Alternate interior angles, as

$l \parallel m$ )

$\therefore \triangle ABC \cong \triangle CDA$ (By ASA congruence rule)

In quadrilateral $ACBD, AC = AD$ and $AB$ bisects $\angle A$ . Show that $\triangle ABC \equiv \triangle ABD$ . What can you say about $BC$ and $BD$

$\triangle ABC$ and

$\triangle ABD$ ,

$AC = AD$ (Given)

$\angle CAB = \angle DAB$ (

$AB$ bisects

$\angle A$ )

$AB = AB$ (Common)

$\therefore \triangle ABC \cong \triangle ABD$ (By SAS congruence rule)

$\therefore BC = BD$ (By CPCT)

$\therefore$ ,

$BC$ and

$BD$ are of equal lengths.

$AB$ is a line segment and $P$ is its mid-point. $D$ and $E$ are points on the same side of $AB$ such that $\angle BAD = \angle ABE$ and $\angle EPA = \angle DPB$ . Show that
(i) $\triangle DAP \cong \triangle EBP$
(ii) $AD = DE$

It is given that

$\angle EPA=\angle DPB$

Now,

$\angle EPA+\angle DPE=\angle DPB+\angle DPE$

Therefore,

$\angle DPA=\angle EPB$

$\Delta EBP$ and

$\Delta DAP$ ,

$\angle EBP = \angle DAP$ (given)

$BP = AP$ (

$P$ is midpoint of

$AB$ )

$\angle EPB=\angle DPA$ [proved above]

By ASA criterion of congruence,

$\Delta EBP \cong \Delta DAP$

ii)

Since

$\Delta EBP \cong \Delta DAP$

$AD = BE$ (using CPCT)

In fig, $ABC$ is a right triangle right angled at $A.BCED, ACFG$ and $ABMN$ are square on the sides $BC, CA$ and $AB$ respectively. Line segment $AX\perp DE$ meets $BC$ at $Y$ . Show that:
(i) $\triangle MBC\cong \triangle ABD$
(ii) $ar(BYXD) = 2ar(MBC)$
(iii) $ar(BYXD) = ar (ABMN)$
(iv) $\triangle FCB\cong \triangle ACE$
(v) $ar(CYXE) = 2ar (FCB)$
(vi) $ar(CYXE) = ar(ACFG)$
(vii) $ar(BCED) = ar(ABMN) + ar(ACFG)$
Result (vii) is the famous Theorem of Pythagoras. You shall learn a simpler
proof of this theorem in Class X.

(i)

$\Delta \ MBC$ and

$\Delta ABD$ , we have

$BC = BD$

[Sides of the square BCED]

$MB = AB$

[Sides of the square ABMN]

$\angle MBC=\angle ABD$

[Since Each

$= 90º + ∠ABC$ ]

Therefore by SAS criterion of congruence, we have

$\Delta MBC\cong \Delta ABD$

(ii)

$\Delta ABD$ and square

$BYXD$ have the same base

$BD$ and lie between the same parallel lines

$BD$ and

$AX$ .
Therefore

$ar(ABD) = \dfrac{1}{2} ar (BYXD)$

But

$ΔMBC≅ΔABD$ [Proved in part (i)]

$\Rightarrow ar(MBC) = ar (ABD)$

Therefore

$ar(MBC) = ar (ABD) = \dfrac{1}{2} ar (BYXD)$

$\Rightarrow ar(BYXD) = 2 ar (MBC).$

(iii)

Square

$ABMN$ and

$\Delta MBC$ have the same base

$MB$ and are between same parallel lines

$MB$ and

$NA$ .
Therefore

$ar(MBC) = \dfrac{1}{2} ar (ABMN)$

$\Rightarrow ar(ABMN) = 2ar (MBC)$

$= ar (BYXD)$ [Using part (ii)]

(iv)

$\Delta$ ACE and

$\Delta BCF$ , we have

$CE = BC$ [Sides of the square BCED]

$AC = CF$ [Sides of the square ACFG]
and

$∠ACE=∠BCF$ [Since Each

$= 90º + ∠BCA$ ]
Therefore by SAS criterion of congruence,

$ΔACE≅ΔBCF$

(v)

$\Delta ACE$ and square

$CYXE$ have the same base

$CE$ and are between same parallel lines

$CE$ and

$AYX$ .
Therefore

$ar(ACE) = \dfrac{1}{2} ar (CYXE)$

$\Rightarrow ar(FCB) = \dfrac{1}{2} ar (CYXE)$ [Since

$ΔACE≅ΔBCF$ , part (iv)]

$\Rightarrow ar(CYXE) =2 ar (FCB)$ .

(vi)

Square

$ACFG$ and

$\triangle BCF$ have the same base

$CF$ and are between same parallel lines

$CF$ and

$AG$ .
Therefore

$ar(BCF) = \dfrac{1}{2} ar(ACFG)$

$\Rightarrow \dfrac{1}{2}ar(CYXE) = \dfrac{1}{2} ar (ACFG)$ [Using part (v)]

$\Rightarrow ar(CYXE) = ar (ACFG)$

(vii)

From part (iii) and (vi) we have

$ar (BYXD) = ar (ABMN)$
and

$ar (CYXE) = ar (ACFG)$

On adding we get

$ar (BYXD) + ar(CYXE) = ar (ABMN) + ar(ACFG) \\ ar (BCED) = ar (ABMN) + ar(ACFG)$

In $\triangle ABC$ and $\triangle DEF, AB = DE, AB \parallel DE, BC = EF$ and $BC \parallel EF$ . Vertices $A, B$ and $C$ are joined to vertices $D, E$ and $F$ respectively. Show that
(i) Quadrilateral $ABED$ is a parallelogram
(ii) Quadrilateral $BEFC$ is a parallelogram
(ii) $AD \parallel CF$ and $AD = CF$
(iv) Quadrilateral $ACFD$ is a parallelogram
(v) $AC = DF$
(vi) $\triangle ABC \cong \triangle DEF$

(i) Consider the quadrilateral $ABED$
We have , $AB = DE$ and $AB \parallel DE$

One pair of opposite sides are equal and parallel. Therefore

$ABED$ is a parallelogram.

(ii) In quadrilateral $BEFC$ , we have
$BC = EF$ and $BC \parallel EF$ . One pair of opposite sides are equal and parallel.therefore , $BEFC$ is a parallelogram.

(iii) $AD = BE$ and $AD \parallel BE$ $\mid$ As $ABED$ is a ||gm ... (1)
and $CF = BE$ and $CF\parallel BE$ $\mid$ As $BEFC$ is a ||gm ... (2)

From (1) and (2), it can be inferred

$AD = CF$ and $AD \parallel CF$

(iv) $AD = CF$ and $AD \parallel CF$

One pair of opposite sides are equal and parallel

$ACFD$ is a parallelogram.

(v) Since $ACFD$ is parallelogram.

$AC = DF$ $\mid$ As Opposite sides of a|| gm $ACFD$

(vi) In triangles

$ABC$ and

$DEF$ , we have

$AB = DE$

$\mid$ (opposite sides of

$ABED$

$BC = EF$

$\mid$ (Opposite sides of

$BEFC$

and

$CA = FD$

$\mid$ Opposite. sides of

$ACFD$

Using SSS criterion of congruence,

$\triangle ABC \cong \triangle DEF$

$ABC$ is an isosceles triangle with $AB = AC$ . Draw $AP \perp BC$ . Show that $\angle B = \angle C$

$\Delta ABP$ and

$\Delta ACP,$

$\angle APB=\angle APC$ (Both equal to

$90^0$ )

$AB = AC$ [

$\because \Delta ABC$ is an isosceles triangle.]

$AP = AP$ (Common Side)

By R.H.S. criterion of congruence,

$\triangle ABP\cong \triangle ACP$

$\Rightarrow \angle B=\angle C$ (CPCT)

$\triangle ABC$ and $\triangle DBC$ are two isosceles triangles on the same base $BC$ and vertices $A$ and $D$ are on the same side of $BC$ . If $AD$ is extended to intersect $BC$ at $P$ , show that
(i) $\triangle ABD \cong \triangle ACD$
(ii) $\triangle ABP \cong \triangle ACP$
(iii) $AP$ bisects $\angle A$ as well as $\triangle D$ .
(iv) $AP$ is the perpendicular bisector of $BC$ .

(i) In

$\triangle ABD$ and

$\triangle ACD$ ,

$AB = AC$ ....(since

$\triangle ABC$ is isosceles)

$AD = AD$ ....(common side)

$BD = DC$ ....(since

$\triangle BDC$ is isosceles)

$ΔABD\cong ΔACD$ .....SSS test of congruence,

$\therefore \angle BAD=\angle CAD$ i.e.

$\angle BAP=\angle PAC$ .....[c.a.c.t]......(i)

(ii) In

$\triangle ABP$ and

$\triangle ACP$ ,

$AB = AC$ ...(since

$\triangle ABC$ is isosceles)

$AP = AP$ ...(common side)

$\angle BAP=\angle PAC$ ....from (i)

$\triangle ABP\cong \triangle ACP$ .... SAS test of congruence

$\therefore BP = PC$ ...[c.s.c.t].....(ii)

$\angle APB=\angle APC$ ....c.a.c.t.

(iii) Since

$\triangle ABD\cong \triangle ACD$

$\angle BAD=\angle CAD$ ....from (i)

So,

$AD$ bisects

$\angle A$

i.e.

$AP$ bisects

$\angle A$ .....(iii)

$\triangle BDP$ and

$\triangle CDP$ ,

$DP = DP$ ...common side

$BP = PC$ ...from (ii)

$BD = CD$ ...(since

$\triangle BDC$ is isosceles)

$\triangle BDP\cong \triangle CDP$ ....SSS test of congruence

$\therefore \angle BDP=\angle CDP$ ....c.a.c.t.

$\therefore$

$DP$ bisects

$\angle D$

So,

$AP$ bisects

$\angle D$ ....(iv)

From (iii) and (iv),

$AP$ bisects

$\angle A$ as well as

$\angle D$ .

(iv) We know that

$\angle APB+ \angle APC= 180^{\circ}$ ....(angles in linear pair)

Also,

$\angle APB=\angle APC$ ...from (ii)

$\therefore \angle APB=\angle APC=\dfrac{180^{\circ}}{2}=90^{\circ}$

$BP = PC$ and

$\angle APB=\angle APC=90^{\circ}$

Hence,

$AP$ is perpendicular bisector of

$BC$ .

$ABC$ is a triangle in which altitudes $BE$ and $CF$ to sides $AC$ and $AB$ are equal (see Fig.). Show that
(i) $\triangle ABE\cong \triangle ACF$
(ii) $AB = AC$ , i.e., $ABC$ is an isosceles triangle.

$\Delta ABE$ and

$\Delta ACF$ ,

$\angle BAE=\angle CAF$ (Common angle)

$\angle AEB=\angle AFC$

$....(\because BE \perp AC$ and

$CF \perp AB)$

$BE = CF$ (Given that altitudes are equal)

By AAS criterion of congruence,

$\Delta ABE \cong \Delta ACF$

Hence,

$AB = AC$ (by CPCT)

$AD$ is an altitude of an isosceles triangle $ABC$ in which $AB = AC$ . Show that
(i) $AD$ bisects $BC$ (ii) $AD$ bisects $\angle A$

$AD$ is the altitude drawn from vertex

$A$ of an isosceles

$\triangle ABC$ to the opposite base

$BC$ so that

$AB = AC$ ,

$\angle ADC=\angle ADB= 90^{\circ}$

Now, in

$\triangle ADB$ and

$\triangle ADC$ , we have

Hypotenuse

$AB =$ Hypotenuse

$AC$ (Given)

$AD = AD$ $$ (Common)

and

$\angle ADC=\angle ADB$ (Each equal to

$90^\circ$ )

Therefore by

$RHS$ criterion of congruence, we have

$\triangle ADB\cong \triangle ADC$

$\Rightarrow BD = DC$ and

$\angle BAD=\angle DAC$ $$ (Since corresponding parts of the congruent triangle are equal.)

Hence,

$AD$ bisects

$BC$ , which proves

$(i)$ and

$AD$ bisects

$\angle A,$ which proves

$(ii)$ .

In given figures two sides $AB$ and $BC$ and median $AM$ of one triangle $ABC$ are respectively equal to sides $PQ$ and $QR$ and median $PN$ of $\triangle PQR$ . Show that:
(i) $\triangle ABM\cong \triangle PQN$
(ii) $\triangle ABC \cong \triangle PQR$

$\triangle ABC$ and

$\triangle PQR$ in which

$AB = PQ, BC = QR$ and

$AM = PN$ .

Since

$AM$ and

$PN$ are median of triangles

$ABC$ and

$PQR$ respectively.

Now,

$BC = QR$

$\mid$ Given

$\Rightarrow \dfrac{1}{2}BC=\dfrac{1}{2}QR$

$\mid$ Median divides opposite sides in two equal parts

$BM = QN$ ... (1)

Now, in

$\triangle ABM$ and

$\triangle PQN$ we have

$AB = PQ$

$\mid$ Given

$BM = QN$

$\mid$ From (i)

and

$AM = PN$

$\mid$ Given

$\therefore$ By SSS criterion of congruence, we have

$\triangle ABM\cong \triangle PQN,$ which proves (i)

$∠B=∠Q$ ... (2)

$\mid$ Since, corresponding parts of the congruent triangle are equal

Now, in

$\triangle ABC$ and

$\triangle PQR$ we have

$AB = PQ$

$\mid$ Given

$∠B=∠Q$

$\mid$ From (2)

$BC = QR$

$\mid$ Given

$\therefore$ by SAS criterion of congruence, we have

$\triangle ABC\cong \triangle PQR$ , which proves (ii)

$BE$ and $CF$ are two equal altitudes of a triangle $ABC$ . Using RHS congruence rule, prove that the triangle $ABC$ is isosceles.

$\Delta BCF$ and

$\Delta CBE$ ,

$\angle BFC=\angle CEB$ (Each

$90º$ )

Hyp.

$BC =$ Hyp.

$BC$ (Common Side)

Side

$FC =$ Side

$EB$ (Given)

$\therefore$ By R.H.S. criterion of congruence, we have

$\Delta BCF\cong \Delta CBE$

$\therefore \angle FBC=\angle ECB$ (CPCT)

$\Delta ABC$ ,

$\angle ABC=\angle ACB$

[

$\because \angle FBC= \angle ECB$ ]

$\therefore AB = AC$ (Converse of isosceles triangle theorem)

$\therefore$

$\Delta ABC$ is an isosceles triangle.

In given figure $ABCD$ is a trapezium in which $AB\parallel CD$ and $AD = BC$ . Show that
(i) $\angle A = \angle B$
(ii) $\angle C = \angle D$
(iii) $\triangle ABC \cong \triangle BAD$
(iv) diagonal $AC =$ diagonal $BD$

Given that ABCD is a trapezium,

$AB\parallel CD$ and

$AD=BC$

To prove: (i)

$\angle A = \angle B$

(ii)

$\angle C= \angle D$

(iii)

$\triangle ABC \cong \triangle BAD$

(iv)

$AC =BD$

Construction: Draw

$CE\parallel AD$ and extend AB to intersect CE at E.

Poof:

$(i)$ As AECD is a parallelogram.

By construction, CE is parallel to AD and AE is parallel to CD,

$\therefore \,\,AD=EC$

But,

$AD=BC$ (Given)

$\therefore \,\,BC=EC$

$\Rightarrow \,\angle 3=\angle 4$ (Angle opposite to equal side are equal)

Now,

$\angle 1+\angle 4={{180}^{0}}$ (consecutive interior angle of parallelogram )

And

$\angle 2+\angle 3={{180}^{0}}$ (linear pair )

$\Rightarrow \angle 1+\angle 4=\angle 2+\angle 3$

$\Rightarrow \angle 1=\angle 2\,\,\,\,\,\,\,\,\,\,\,\,\because \angle 3=\angle 4$ (so get cancelled with each other )

$\Rightarrow \,\angle A=\angle B$

$(ii)$

$AB\parallel CD$

$\therefore \ \angle A +\angle D=\angle B+\angle C=180^o$ [Angles on the same side of transversal]

But,

$\angle A=\angle B$

Hence,

$\angle C=\angle D$

$(iii)$ In

$\triangle ABC$ and

$\triangle BAD$ , we have

$BC = AD$ (Given)

$AB = BA$ (Common)

$\angle A = \angle B$ (Proved)

Hence, by SAS congruence criterion,

$\triangle ABC \cong \triangle BAD$

$(iv)$ We have proved that,

$\triangle ABC \cong \triangle BAD$

$\therefore \ AC = BD$ (CPCT)

Hence, all the four required results have been proved.

2105599_463886_ans_2d93e317545e48c0bc67d3da3736bc03.png

Show that the sum of the three altitudes of a triangle is less than the sum of its three sides.

Let

$ABC$ be the triangle with sides

$BC, AC, AB$ opposite angles

$A,B,C$ being in length equal to

$a,b,c$ respectively.

Draw

$AD, BE, CF$ perpendicular from

$A,B,C$ to opposite sides meeting sides

$BC, AC, AB$ at

$D,E, F$ respectively.

Now perpendicular

$AD^2=AC^2-CD^2$

$\Rightarrow AD^2<AC^2$

$\Rightarrow AD<AC$

$\Rightarrow AD<b$ .... (1)

Also,

$BE^2=AB^2-AE^2$

$\Rightarrow BE^2<AB^2$

$\Rightarrow BE<AB$

$\Rightarrow BE<c$ ..... (2)

Likewise

$CF^2=BC^2-BF^2$

$\Rightarrow CF^2<BC^2$

$\Rightarrow CF<BC$

$\Rightarrow CF<a$ ..... (3)

Adding inequalities, (1), (2) and (3), we get

$AD+BE+CF<a+b+c$

Hence Proved.

In $\Delta ABC \,and \, \Delta PQR$ , AB = PQ, BC = QR; CB and RQ are extended to X and Y respectively. $\angle ABX= \angle PQY$ . Prove that $\Delta ABC = \Delta PQR$ .

$\Delta ABC$ and

$\Delta PQR$

$AB=PQ$

And

$BC=QR$

$CB$ And

$RQ$ extended to

$X$ and

$Y$

$\angle ABX=\angle PQY$

$\angle ABC=180^{o}-\angle ABX=180^{o}\angle PQY$

$\angle PQR=180^{o}-\angle PQY=\angle ABC$

We get in both triangles

$AB=PQ$

$BC=QR$

And

$\angle ABC=\angle PQR$

Then both triangle are congruent

i.e.

$\Delta ABC=\Delta PQR$

In the given figure, AB = CF, EF = BD $\angle AFE = \angle DBC$ . Prove that $\Delta AFE = \Delta CBD$

Given in figure

$AB=CF$ ........... (1)

$EF=BD$ ........... (2)

$\angle AFE=\angle DBC$

Then

$AF=AB +BF$

And

$BC=BF+CF$

As per (1) and (2) we get

$BC=BF+AB=AF$

Thus in

$\Delta AFE$ and

$\Delta BDC$

$BC=AF$

$EF=BD$

$\angle AFE=\angle DBC$

$\therefore \Delta AFE=\Delta CBD$ .......... [By SAS rule of congruence]

In the figure, $\angle BCD = \angle ADC$ and $\angle ACB = \angle BDA$ . Prove that AD = BC and $\angle A = \angle B$ .

We have

$\angle 1 = \angle 2$ and

$\angle 3 = \angle 4$

$\Rightarrow \angle 1 + \angle 3 = \angle 2 + \angle 4$

$\Rightarrow \angle ACD = \angle BDC$ .
Thus in triangles ACD and BDC, we have,

$\angle ADC = \angle BCD$ (given);
CD = CD (common);

$\angle ACD = \angle BDC$ (proved).

By ASA condition

$\triangle ACD \cong \triangle BDC$ . Therefore

$AD = BC$ and

$\angle A = \angle B$ .

In a $\triangle ABC, AB = AC$ and $\angle A = 50^o$ . Find $\angle B$ and $\angle C$ .

We can see that,

$\angle B$ =

$\angle C$ (Since AB = AC)

$\angle B$ +

$\angle C$ +

$\angle A$ =

$180^\circ$ [sum of all the angles of a triangle is

$180^o$ ]

$\Rightarrow \angle B$ +

$\angle C$ =

$180^\circ-50^\circ$

$\Rightarrow$

$2\angle C$ =

$180^\circ-50^\circ$

$\Rightarrow \angle C$ =

$\frac {130^\circ}{2}$

$\Rightarrow \angle B$ =

$65^\circ$ and

$\angle C$ =

$65^\circ$

In a $\triangle ABC,$ we have $AB = 4 \ cm$ , $BC = 5.6 \ cm$ and $CA = 7.6 \ cm$ . Write the angles of the triangle in ascending order of measures.

Axiom

$1:$ Angle opposite to the largest side is largest and vice-versa.

Axiom

$2:$ Angle opposite to smallest side is smallest and vice-versa.

By using axiom

$1,$

From given data

$AC = 7.6\ cm$ is largest, the angle opposite to

$AC$ is

$\angle B.$

$\therefore$ Largest angle in

$\triangle ABC$ is

$\angle B.$

By using axiom

$2,$

Also,

$AB=4\ cm$ is smallest, the angle opposite to

$AB$ is

$\angle C.$

$\therefore$ Smallest angle in

$\triangle ABC$ is

$\angle C.$

Hence, the angles in ascending order will be,

$\angle C<\angle A<\angle B$

In the given figure, find the value of $x$

$\textbf{Step -1: Apply the first property mentioned.}$

$\text{Given figure is, }$

$\text{We know that, BD = DC = AD}$

$\angle\text{ABD}=50^{\circ}$

$\text{In triangle ABD, BD = AD}$

$\therefore\angle\text{BAD}=\angle\text{ABD}=50^{\circ}$

$\text{Also, }\angle\text{ABD}+\angle\text{BAD}=\text{External }\angle\text{ADC}\Rightarrow\text{External }\angle\text{ADC}=50^\circ+50^\circ=100^\circ$

$\textbf{Step -2: Finding the value of x }$

$\text{In triangle ADC, AD = DC}$

$\angle\text{ACD}=\angle\text{DAC}=x$

$\text{According to angle property, the sum of the angles of a triangle is }180^\circ.$

$\angle\text{ACD}+\angle\text{DAC}+\angle\text{ADC}=180^\circ$

$\Rightarrow{x}+x+100^\circ=180^\circ$

$\Rightarrow2x=180^\circ-100^\circ=80^\circ$

$\Rightarrow{x}=\dfrac{80^\circ}{2}=40^\circ$

$\textbf{Hence, the value of }\mathbf{x}\textbf{ is }\mathbf{50^\circ}.$

In given figure diagonal AC of a quadrilateral, ABCD bisects the angles $\angle A$ and $\angle C$ . Prove that AB = AD and CB = CD.

Since diagonal AC bisects the angles

$\angle A$ and

$\angle C$ ,

we have

$\angle BAC = \angle DAC$ and

$\angle BCA = \angle DCA$ .

In triangles ABC and ADC,

we have

$\angle BAC = \angle DAC$ (given);

$\angle BCA = \angle DCA$ (given);

AC = AC (common side).

So, by ASA postulate, we have

$\triangle BAC \cong \triangle DAC$

$\Rightarrow$ BA = AD and CB = CD (Corresponding parts of congruent triangle).

In the figure O is the midpoint of AB and CD. Prove that
(i) $\triangle AOC \cong \triangle BOD$ ; (ii) $AC = BD$ .

In triangles AOC and BOD, we have
AO = BO (O, the midpoint of AB);

$\angle AOC = \angle BOD$ , (vertically opposite angles);

$CO = OD$ , (O, the midpoint of CD)
So by SAS postulate we have

$\triangle AOC \cong \triangle BOD$ .
Hence, AC = BD, as they are corresponding parts of congruent triangles.

In the figure, find the value of $x$ .

$\angle A$ =

$\angle D$ =

$30^\circ$
In

$\Delta ACD, \angle A + \angle C + \angle D = 180^\circ$
(Sum of all angles of

$\Delta$ is

$180^\circ$ )

$30^\circ+ \angle C + 30^\circ = 180^\circ$

$\angle C = 180^\circ - 60^\circ$

$\angle C = 120^\circ$

$\angle ACB +\angle ACD = 180^\circ$ (Linear pair)

$\angle ACB = 180^\circ - 120^\circ= 60^\circ$
In

$\angle ABC$ ,

$\angle A + \angle B + \angle C = 180^\circ$
(Sum of all angles of

$\Delta$ is

$180^\circ$ )

$65^\circ+ x+ 60^\circ$ =

$180^\circ$

$x = 180^\circ - 125^\circ$

$x= 55^\circ$

In a triangle ABC, AB = AC and the bisectors of angles B and C intersect at O. Prove that BO = CO and AO is the bisector of angle $\angle BAC$ .

Since the angles opposite to equal sides are equal,

$\therefore AB = AC$

$\Rightarrow \angle C = \angle B$

$\Rightarrow \dfrac{\angle B}{2} = \dfrac{\angle C}{2}$ .

Since

$BO$ and

$CO$ are bisectors of

$\angle B$ and

$\angle C$ , we also have

$\angle ABO = \dfrac{ \angle B}{2}$ and

$\angle ACO = \dfrac{\angle C}{2}$ .

$\angle ABO = \dfrac{\angle B}{2} = \dfrac{\angle C}{2} = \angle ACO$ .

Consider

$\triangle BCO$ :

$\angle OBC = \angle OCB$

$\Rightarrow BO = CO$ ....... [Sides opposite to equal angles are equal]

Finally, consider triangles

$ABO$ and

$ACO$ .

$BA = CA$ ...... (given);

$BO = CO$ ...... (proved);

$\angle ABO = \angle ACO$ (proved).

Hence, by

$\text{S.A.S}$ postulate

$\triangle ABO \cong \triangle ACO$

$\Rightarrow \angle BAO = \angle CAO \Rightarrow AO$ bisects

$\angle A$ .

In the figure AB = AC and DB = DC. Prove that $\angle ABD = \angle ACD$ .

$\triangle ABC$ , we have

$AB = AC \Rightarrow \angle ABC = \angle ACB$

(angles opposite to equal sides.)

Again in

$\triangle DBC$ , we have

$DB = DC$ ......... (given)

$\Rightarrow \angle DBC = \angle DCB$ (angles opposite to equal sides).

Hence we obtain

$\angle ABC - \angle DBC = \angle ACB - \angle DCB$ .

This gives

$\angle ABD = \angle ACD$ .

$[hence \quad proved]$

In given figure the altitudes AD, BE and CF of triangle ABC are equal. Prove that ABC is an equilateral triangle.

In right-angle triangles BCE and CBF, we have,

BC = BC (common hypotenuse);

BE = CF (given).

Hence BCF and CBF are congruent, by RHS theorem. Comparing the triangles, we get

$\angle B = \angle C$ .

This implies that

AC = AB (sides opposite to equal angles).

Similarly,

$AD = BE \Rightarrow \angle B = \angle A$

$\Rightarrow AC = BC$

Together, we get

$AB = BC = AC$ or

$\triangle ABC$ is equilateral.

$[hence \quad proved]$

If $\triangle ABC \cong \triangle NMO$ , name the congruent sides and angles.

If we are taking

$\triangle ABC \cong \triangle NMO$

Then,

(i) By

$c.p.c.t.$ , the congruent angles are as below

$\angle A = \angle N; \angle B=\angle M; \angle C=\angle O$

(ii) By

$c.p.c.t$ , Congruent sides are as below

$AB=NM; BC=MO$ and

$AC=NO$

Prove that the base angles of an isosceles trapezium are equal

Given:

$\overline {AB}=\overline {CD}$

To Prove:

$\angle A=\angle D$ .

Proof:-

Let

$ABCD$ is a trapezium with

$AB=CD.$ Drop perpendiculars from

$B$ and

$C$ as shown in the above figure. So, we have

$BE$ and

$CF$ perpendicular to

$AD.$

Now, in

$\triangle ABE$ and

$\triangle CFD$ ,

$\overline {BE} = \overline {CF}\quad$ (altitudes of trapezium)

$\overline {AB}=\overline {CD}\quad$ (Given)

$\angle AEB = \angle DFC\quad$ (right angle)

From

$R.H.S$ congruence,

$\triangle ABE \cong \triangle CFD$

$\angle A =\angle D\quad \dots \left[CPCT\right]$

Hence proven base angles of an isosceles trapezium are equal.

Which minimum measurements do you require to check if the given figures are congruent:
Two rectangles

In the case of a rectangle, the length and breadth should be same. All angles of a rectangle, square are of

$90^o$ . That is a common check.

What are the minimum measurements required for two rhombuses to be congruent?

In the case of a rhombus, two figures are similar by definition if all the corresponding sides are proportional and all their corresponding angles are congruent.

A rhombus being a parallelogram, its opposite angles are congruent and its consecutive angles are supplementary.

Suppose ABC is an isosceles triangle such that AB = AC and AD is the altitude from A on BC. Prove that (i) AD bisects $\angle A$ , (ii) AD bisects BC.

We have to show that

$\angle BAD = \angle CAD$ and BD = DC.
In right triangles ADB and ADC, we have
AB = AC (given);
AD = AD (common side).
So by RHS congruency of triangles, we have

$\triangle ABD \cong \triangle ACD$ . Hence,

$\angle BAD = \angle CAD$ and

$BD = DC$ .

State whether the following triangles are congruent or not? Give reasons for your answer

Consider

$\triangle ABC$ ,

By angle sum property,sum of all angles

$=180^o$

$\angle A+\angle B+\angle C=180^o$

$\Rightarrow \angle A=180^o - 50^o-70^o=60^o$

Consider

$\triangle DEF$

$\angle F+\angle E+\angle D=180^o$

$\Rightarrow \angle E=180^o -70^o-60^o=50^o$

$\therefore \angle A=\angle D=60^o$

$\angle B=\angle E=50^o$

$\angle C=\angle F=70^o$

and

$BC=FE$

$\therefore \triangle ABC \cong \triangle FED$ ...... [ASA Criterion]

In the figure, it is given that AB = CD and AD = BC. Prove that triangles ADC and CBA are congruent.

In triangles

$ADC$ and

$CBA$ , we have

$AB = CD$ ......... (given);

$AD = BC$ ......... (given);

$AC = AC$ ......... (common side).

$\text{S.S.S}$ congruency condition,

$\triangle ADC \cong \triangle CBA$

$[hence \quad proved]$

State whether the following triangles are congruent or not? Give reasons for your answer.

Consider

$\triangle LMN$ and

$\triangle RTS$

$LN=RS=4cm$

$MN=ST=3cm$

$\therefore \angle LMN=\angle RTS=80^o$

$\therefore \triangle LMN \cong \triangle RTS$ [by SAS Criteria]

In the adjoining figure, AB = CD and AD = BC. Show that $\angle 1 = \angle 2$

In triangles

$ABD$ and

$CDB,$

We have

$AB = CD$ (Given)

$AD = BC$ (Given)

$BD = DB$ (common side)

Hence,

$\triangle ABD \cong \triangle CDB$ , by

$\text{SSS}$ postulate.

Comparing the angles of the congruent triangles

we get

$\angle ADB=\angle CBD$

$\therefore \ \angle 1 = \angle 2$

$\triangle ABC$ and $\triangle DBC$ are two isosceles triangles on the same base $BC$ (see figure). Show that $\angle ABD =\angle ACD$

Given

$\Delta ABC$ and

$\Delta DBC$ are two isosceles triangles on the same base BC

In isosceles

$\Delta ABC$

$AB=AC$

Then

$\angle ABC=\angle ACB$ ................................(1)

In isosceles

$\Delta BDC$

$BD=DC$

Then

$\angle CBD=\angle BCD$ ..................................(2)

Add (1) and (2) we get

$\angle ABC+\angle CBD=\angle ACB+\angle BCD$

But

$\Delta ABC$ and

$\Delta DBC$ on same base

$\therefore \angle ABD=\angle ACD$

$l$ and $m$ are two parallel lines intersected by another pair of parallel lines $p$ and $q$ . Show that $\triangle ABC\cong \triangle CDA$

Its given that

$l||m$ and

$p||q$ . We have to prove that

$\triangle ABC \cong \triangle CDA$

Taking

$l||m$ and

$AC$ is the transversal,

$\angle ACB=\angle CAD$ [alternate angle]....(i)

Considering

$p||q$ and

$AC$ is the transversal,

$\angle BAC=\angle DCA$ [alternate angles].....(ii)

$\therefore$ In

$\triangle ABC$ and

$\triangle CDA$ ,

$\angle ACB=\angle CAD$ [alternate angles]

$AC=AC$ [common base]

$\angle BAC=\angle DCA$ [alternate angles]

$\therefore \triangle ABC \cong \triangle CDA$ [ASA Congruence Rule]

$BE$ and $CF$ are two equal altitudes of a triangle $ABC$ . Using $RHS$ congruence rule, prove that the triangle $ABC$ is isosceles

$\textbf{Step 1: Showing that the 2 angles of }\mathbf{\triangle ABC}\textbf{ are equal}$

$\text{In }\triangle BCF\text{ and }\triangle CBE$

$BC=BC\text{ (Common Side)}$

$BE=CF\text{ (Given)}$

$\therefore\triangle BCF\cong\triangle CBE\text{ (RHS criterion of congruence)}$

$\implies \angle FBC=\angle ECB\text{ (By CPCT)}$

$\textbf{Step 2: Proving triangle ABC is isosceles}$

$\angle FBC=\angle ECB$

$\implies \angle FBC=\angle ECB$

$\therefore AB=AC\text{ (Converse of isosceles triangle theorem)}$

$\therefore \triangle ABC \text{ is isosceles}$

$\textbf{Hence proved that triangle ABC is isosceles}$

In the adjacent figure $\triangle ABC, D$ is the midpoint of $BC$ . $DE\perp AB, DF\perp AC$ and $DE = DF$ . Show that $\triangle BED \cong \triangle CFD$

From

$\triangle BED$ and

$\triangle CFD$

$BD= CD$ (D is a midpoit of BC)

$DE =DF$ (Given)

$\angle DEB = \angle DFC$ (

$DE \bot AB$ and

$DF \bot AC$ )

Thus,

$\triangle BED \cong \triangle CFD$ (By RHS)

1025802_569799_ans_a26986e6b14d4464bf0f3f54be306ede.png

In the adjacent figure $\triangle ABC$ is isosceles as $\overline {AB} = \overline {AC}, \overline {BA}$ and $\overline {CA}$ are produced to $Q$ and $P$ such that $\overline {AQ} = \overline {AP}$ . Show that $\overline {PB} = \overline {QC}$

Consider

$\triangle PAB$ and

$\triangle QAC$

$\overline {PA}=\overline {QA}$ ..... [given]

$\overline {AB}=\overline {AC}$ ...... [given]

$\angle PAB=\angle QAC$ ...... [vertically opposite angles]

$\therefore$ By SAS Congruency Rule,

$\triangle PAB \cong \triangle QAC$

So,

$\overline{PB}=\overline{QC}$ ....... [By CPCT]

In given figure $AD$ is an altitude of an isosceles triangle $ABC$ in which $AB = AC$ .
Show that, $AD$ bisects $\angle A$

$\triangle ABC$ is an isosceles triangle,

$AB=AC$

$AD$ is the altitude,

$\angle ADC=\angle ADB=90^o$

$\triangle ADB$ and

$\triangle ADC$ ,

$\angle ADC=\angle ADB=90^o$

$AB=AC$ and

$AD=AD$ [common]

$\therefore \triangle ADB \cong \triangle ADC$ [RHS Congruency]

Hence by CPCT,

$\Rightarrow BD=DC$

$\therefore \angle BAD=\angle DAC$

i.e,

$AD$ bisects

$\angle A$

$[hence \quad proved]$

In the adjacent figure $ABCD$ is a square and $\triangle APB$ is an equilateral triangle. Prove that $\triangle APD\cong \triangle BPC$

Given that

$APB$ is an equilateral triangle, so the sides of an equilateral triangle are equal.

$AP = BP$ ...... (1)

and

$ABCD$ is a square. All the sides of a square are equal.

$AD = BC$ .....(2)

and

$\angle PAD=\angle DAB - \angle PAB=90^{0}-60^{0}=30^{0}$

$\angle PBC=\angle ABC - \angle PBA=90^{0}-60^{0}=30^{0}$

$\therefore \angle DAP=\angle BPC$ ......(3)

So, from the

$S.A.S.$ congruency,

$\therefore \Delta APD\cong \Delta BPC$

ABC is a right angled triangle in which $\angle A = 90^{\circ}$ and AB = AC. Show that $\angle B = \angle C$ .

Given:

$ABC$ is a right-angled triangle with

$\angle A =90^o$ and

$AB=AC$

Draw AD perpendicular from point A on line BC

$\Delta ADC$ and

$\Delta ADB$

$AB=AC$ ...Given

$AD=AD$ ...Common side

$\angle ADC=\angle ADB = 90^o$

Hence, by RHS congruence criterion,

$\Delta ADC\cong ADB$

$\therefore \ \angle B=\angle C$ ....CACT

$\quad [hence \ proved]$

$\triangle ABC$ is an isosceles triangle in which $AB = AC$ . Show that $\angle B = \angle C$

We have to draw

$AD$ perpendicular to

$BC$ .

Since

$\triangle ABC$ is isosceles,

$AB=AC.......(1)$

Now,

$AD$ perpendicular to

$BC$

$\therefore \angle ADB=\angle ADC=90^o.......(2)$

$\triangle ABD$ and

$\triangle ACD$ ,

$\angle ADB=\angle ADC=90^o$ [From

$(2)$ ]

$AB=AC$ [From

$(1)$ ]

$AD=AD$ [common]

$\triangle ABD \cong \triangle ACD$ [

$RHS$ Congruence Rule]

$\therefore \angle B=\angle C$ [by

$CPCT$ ]

Line-segment $AB$ is parallel to another line-segment $CD. O$ is the mid-point of $AD$ .
Show that $O$ is also the midpoint of $BC$ .

Given: AB is parallel to another line segment CD.

O is the midpoint OF AD

$\Delta AOB$ and

$\Delta DOC$

$\angle AOB=\angle COD$ ...(Vertically opposite angle )

$\angle BAO=\angle CDO$ ...(Given AB parallel to DC and AD meet both lines so alternate angles are equal)

$AO=OD$ ....(O is the midpoint of AD )

$\Delta AOB\cong \Delta DOC$ ...ASA test

So,

$BO=CO$

Then, O is the midpoint of BC.

If two sides of a triangle measure $4cm$ and $6cm$ find all possible measurement (positive Integers) of the third side. How many distinct triangles can be obtained?

We know that the sum of two sides should be less than the third side.

So, let the length of the third side be

$x$ cm.

$4+x>6= x>2$

$x<4+6 = x<10$

Hence,

$x=3,4,5,6,7,8,9$

In $\triangle ABC$ , the bisector $AD$ of $A$ is perpendicular to side $BC$ . Show that $AB = AC$ and $\triangle ABC$ is isosceles.

Given in

$\Delta ABC$ , the bisector AD of

$\angle A$ is perpendicular to side BC.

$\Delta BAD$ and

$\Delta CAD$ ,

$\angle BAD=\angle CAD$ ....Given

$\angle ADB=\angle ADC=90^{0}$ .........(AD is perpendicular to side BC)

$AD=AD$ .............(Common side)

$\therefore \Delta BAD\cong \Delta CAD$ ........[ASA test]

$\therefore AB=AC$ ...........[CSCT]

By converse of isosceles triangle theorem,

$\therefore\triangle ABC$ is isosceles.

$[hence \, proved]$

In adjacent figure, $PR > PQ$ and $PS$ bisects $\angle QPR$ . Prove that $\angle PSR > \angle PSQ$

In the given triangle

$PQR$ ,

$PR > PQ$ and

$PS$ bisects

$\angle QPR$

$\Delta PQR$

$PR > PQ$ ....(Given )

$\therefore \angle PQR > \angle PRQ$ (Angle in the opposite of the longer side is greater) .............................(1)

$\Delta PQS$ and

$\Delta RPS$

$\angle PQR= \angle PQS=180-(\angle PSQ+\angle QPS)$

$\angle PRQ=\angle PRS=180-(\angle PSR+\angle RPS)$

Given

$PS$ is bisector of

$\angle QPR$

Then

$\angle QPS=\angle RPS$ ...........................(2)

$\angle PQR > \angle PRQ$ (As per (1))

$\therefore 180-(\angle PSQ+\angle QPS)> 180-(\angle PSR+\angle RPS)$ (

$\angle QPS=\angle PRQ)$

$\Rightarrow \angle PSR > \angle PSQ$

$[hence \, proved]$

In the figure, $AB\parallel DC$ and $AD\parallel BC$ Show that $\triangle ABC \cong \triangle CDA$ .

Consider

$\triangle ABC$ and

$\triangle CDA$

$\angle BAC = \angle DCA$ ...(Alternate interior angles)

$AC = CA$ ...(Common side)

$\angle BCA = \angle DAC$ (Alternate interior angles)

$\triangle ABC\cong \triangle CDA$ ...(By ASA congruency)

Show that in a right triangle, the hypotenuse is the longest side

Let

$ABC$ be the right angled triangle in which

$\angle B=90^o$

$\therefore \angle A+\angle C=90^o$

Since sum of all the angles of a triangle

$=180^o$ [angle sum property]

$\angle B=\angle A+\angle C$

$\Rightarrow \angle B > \angle A$

and

$\angle B > \angle C$

$\Rightarrow AC > BC$

$\therefore$ Side opposite to greater angle is longer and

$AC > AB$

$\therefore AC$ is the longest side, i.e hypotenuse is the longest side.

In adjacent figure, $\angle B < \angle A$ and $\angle C < \angle D$ . Show that $AD < BC$

$\triangle AOB$ ,

$\angle B < \angle A$

It is known that in a triangle side opposite to the greater angle is longer.

$\therefore AO < BO$ .............

$(1)$

$\triangle COD$ ,

$\angle C < \angle D \Rightarrow OD < OC$ ...........

$(2)$

Add

$(1)$ and

$(2)$ ,

$\Rightarrow AO+OD < OB+OC$

$\therefore AD<BC$

In the given figure, $AD$ is perpendicular to $BC$ and $EF \parallel BC$ , if $\angle EAB = \angle FAC$ , show that triangles $ABD$ and $ACD$ are congruent.
Also, find the values of $x$ and $y$ if $AB = 2x + 3, AC = 3y + 1, BD = x$ and $DC = y + 1$

$AD$ is perpendicular to

$EF$

$\Rightarrow \angle EAD = \angle FAD = 90^{\circ}$

$\angle EAB = \angle FAC$ (given)

$\Rightarrow \angle EAD - \angle EAB = \angle FAD - \angle FAC$

$\Rightarrow \angle BAD = \angle CAD$
In

$\triangle ABD$ and

$\triangle ACD$

$\angle BAD = \angle CAD$ [proved above]

$\angle ADB = \angle ADC = 90^{\circ}$ [Given

$AD$ is perpendicular on

$BC$ ]
and

$AD = AD$

$\triangle ABD \cong \triangle ACD$ [By ASA]
Hence proved.

$\angle ABD = \angle ACD \Rightarrow AB = AC$ and

$BD = CD$ [By C.P.C.T]

$\Rightarrow 2x + 3 = 3y + 1$ and

$x = y + 1$

$\Rightarrow 2x - 3y = -2$ and

$x - y = 1$
Substituting

$2(1 + y) - 3y = -2$ Substituting

$y = 4$ in

$x = 1 + y$

$x = 1 + y\ 2 + 2y - 3y = -2\ x = 1 + 4$

$-y = -2 - 2\ x = 5$

$-y = -4$

In the given figure, $AL\parallel DC, E$ is mid point of $BC$ . Show that $\triangle EBL\cong \triangle ECD$

Consider

$\triangle EBL$ and

$\triangle ECD$

$\angle BEL = \angle CED$ (vertically opposite angles)

$BE = CE$ (since

$E$ is mid point of

$BC$ )

$\angle EBL = \angle ECD$ (alternate interior angles)

$\triangle EDL \cong \triangle ECD$ (by ASA congruency)

In the given figure, $AB$ and $CD$ intersect at $'O', OA = OB$ and $OD = OC$ . Show that $\triangle AOD \cong \triangle BOC$ .

Given: AB and CD intersect at point O.

$OA=OB$ and

$OD=OC$

$\Delta AOD$ and

$\Delta BOC$ ,

$OA=OB$ ....(Given )

$OD=OC$ ...(Given )

$\angle AOD=\angle BOC$ ...(Vertically opposite angles )

$\therefore \Delta AOD \cong \Delta BOC$ ....(SAS test)

$[hence \, proved]$

In the figure , $PQSR$ is a parallelogram.
$PQ=4.3cm$ and $QR=2.5cm$ . Is $\triangle PQR\equiv \triangle PSR$ ?

Consider

$\triangle PQR$ and

$\triangle PSR$ . Here

$PQ=SR=4.3cm$
and

$PR=QS=2.5cm$ .

$PR=PR$ (common side)

$\therefore \triangle PQR\equiv \triangle RSP$

$\therefore \triangle PQR$ and

$\triangle RSP$ are of different order.

In $\triangle^{s}ABC$ and $DEF, AB\parallel DE; BC = EF$ and $BC\parallel EF$ . Vertices, $A, B$ and $C$ are joined to vertices $D, E$ and $F$ respectively (see figure). Show that
$\triangle ABC \cong \triangle DEF$

Given:

$\Delta ABC$ and

$\Delta DEF,AB\left | \right |DE; AB=DE;BC=EF;BC\left | \right |EF$

To show:

$\Delta ABC\cong\Delta DEF$

Proof:

As we know that

$ABCD$ is a parallelogram, so,

$AD||BE$ and

$AD=BE$

As we know that

$BEFC$ is a parallelogram, so,

$BE\left | \right |CF$ and

$BE=CE.$

Now,if

$AD=BE$ and

$BE = CF$ so,

$AD || CF$

Also, if

$AD=BE$ and

$BE=CF$ so,

$AD=CF$ .

This, implies one pair of opposite sides are equal and parallel to each other.

Therefore,

$ADFC$ is a parallelogram.

Therefore,

$AC=DF$

In triangles

$ABC$ and

$DEF$ ,

$AB = DE$ (opposites sides of parallelogram)

$BC = EF$ (opposites sides of parallelogram)

$AC = DF$ (proven above)

So, by SSS criterion of congruency,triangle

$ABC$ is congruent to triangle

$DEF$ .

In the figure, ABCD is a parallelogram is produced to E such that $AB = BE$ . AD produced to F such that $AD = DF$ . Show that $\triangle FCD \cong \triangle CBE$ .

In given figure

$ABCD$ is a parallelogram

$\Rightarrow \angle CBA=\angle CDA$ ....... (opposite angle of parallelogram)

$\therefore 180^{o}-\angle CBA=180^{o}-\angle CDA$

$\therefore \angle CBE=\angle FDC$ .......... (1)

Now,

$AB=DC$ ..... (opposite sides of parallelogram)

$AB=BE$ ...... (given)

$\therefore BE=DC$ ................(2)

Also,

$BC=AD$ ...... (opposite sides of parallelogram)

and

$BC=DF$ .......... (given)

$\therefore BC=DF$ ....................(3)

From

$(1),(2)$ and

$(3)$

$\bigtriangleup CBE \cong \bigtriangleup FCD$ ......... [By SAS congruence rule]

In figure, BO bisects $\angle ABC$ of $\triangle ABC$ . P is any point on BO. Prove that the perpendicular drawn from P to BA and BC are equal.

In the given figure

$BO$ is the angle bisector of

$\angle ABC$

Then

$\angle DBP=\angle DBE=\frac{1}{2}\angle ABC$

Perpendicular drawn from

$P$ to

$AB$ and

$BC$

$\triangle PBD$ and

$\triangle PBE$

$\therefore \angle PDB=\angle PEB=90^{o}$

$BO=BO$ ...... [common side]

$\angle DBP=\angle BDE$

$\therefore\ \bigtriangleup BDP\cong \bigtriangleup BEP$ ....... [By ASA rule]

$\therefore PD=PE$

Then, the perpendicular draw from P to BA and BC are equal.

In given figure $D$ is a point on side $BC$ of $\triangle ABC$ such that $AD = AC$ . Show that $AB > AD$

$\Delta DAC$ ,

$AD = AC$ (Given)

So,

$\angle ADC = \angle ACD$ (Angles opposite to equal sides)

Now,

$\angle ADC$ is an exterior angle for

$\triangle ABD$ .

So,

$\angle ADC > \angle ABD$

$\angle ACD > \angle ABD$

$\angle ACB > \angle ABC$

$AB > AC$ (Side opposite to larger angle in

$\triangle ABC$ )

$AB > AD$

$(AD = AC)$

$[hence \, proved]$

Prove that, if the areas of two similar triangles are equal, then they are congruent.

$\textbf{Step - 1 : Express ratio of areas in terms of sides.}$

$\text{Let the triangles are T and T'.}$

$\text{Let sides of triangle T be a,b and c.}$

$\text{and sides of triangle T' be d,e and f.}$

$\text{Given that, their areas are same,}$

$\therefore \text{Area of T}=\text{Area of T'}$

$\Rightarrow \dfrac{\text{Area of T}}{\text{Area of T'}}=1.......(1)$

$\text{We know, the ratio of areas of 2 similar triangles is equal to the,}$

$\text{square of ratio of their corresponding sides.}$

$\therefore \dfrac{\text{Area of T}}{\text{Area of T'}}=\dfrac{a^2}{d^2}=\dfrac{b^2}{e^2}=\dfrac{c^2}{f^2}$

$\Rightarrow1=\dfrac{a^2}{d^2}=\dfrac{b^2}{e^2}=\dfrac{c^2}{f^2}$

$\textbf{[From (1)]}$

$\Rightarrow \dfrac{a}{d}=\dfrac{b}{e}=\dfrac{c}{f}=1$

$\Rightarrow a=d, b=e, c=f$

$\text{Since, 3 sides of triangle T are equal to 3 sides of triangle T'}$

$\text{Therefore triangle T and triangle T' are congruent by SSS congruence theorem.}$

$\textbf{Hence proved that, if the area of two similar triangles are equal, then they are congruent.}$

$ABCD$ is a parallelogram. $AP$ and $CQ$ are perpendicular lines drawn from vertices $A$ and $C$ respectively on diagonal $BD$ (see figure). Show that
$\triangle APB\cong \triangle CQD$

Given :

$ABCD$ is a parallelogram.

$AP$ and

$CQ$ are perpendicular lines drawn from vertices A and C respectively on diagonal

$BD$ .

To show :

$\Delta APB\cong \Delta CQD$

Proof:

In triangle

$APB$ and triangle

$CQD$ .

$AB= CD$ (Opposites sides of parallelogram)

$\angle ABP=\angle CDQ$ ( Alternate interior angles)

$\angle APB=\angle CQD= 90^{\circ}$ (Given )

Therefore, by

$AAS$ criterion of congruency, triangle

$APB$ is congruent to triangle

$CQD$ .

From the given figures, state whether the given pairs of triangles are congruent by SSS axiom

Compare the sides of the

$\triangle {PQR}$ and

$\triangle {XYZ}$

$PQ=XY=5cm$ ,

$QR=YZ=4.5cm$ and

$RP=ZX=3cm$
If we suppose

$\triangle {PQR}$ on

$\triangle {XYZ}$

$P$ lies on

$X$ ,

$Q$ lies on

$Y$ ,

$R$ lies on

$Z$ and

$\triangle {PQR}$ covers

$\triangle {XYZ}$ exactly.

$\quad \therefore \triangle PQR\equiv \triangle XYZ$ [by SSS axiom]

Decide whether the SSS congruence is true with the following figures. Give reasons

SSS congruence is not true.

Because only two sides of

$\triangle LSD ,\ \triangle ADL$ are equal which are

$SD=AL$ and one common side

$LD.$

Whereas

$LS=2.5, AD=3$

What additional information do you need to conclude that the two triangles given here under are congruent using SAS rule?(When $\,\, HG = TR$ )

SAS Congruence rule :Two triangle are congruent if two sides and the included angle of one triangle are equal to the two sides and the included angle of the other triangle.

$\triangle{GHJ}$ and

$\triangle{RTS},HJ = TS$ ( side ) ---

$( 1 )$

$\angle{GHJ} =\angle{STR}$ ( included angle )--

$( 2 )$

$HG = TR$ ( side )-----

$( GIVEN )$

$\therefore ,\triangle{GHJ}$ Congruent to

$\triangle{RTS}$

Information

$( 1 )$ and

$( 2 )$ required to prove ( SAS ) congruence.

For the following congruent triangles, find the pairs of corresponding angles.

$\triangle TPQ$ and

$\triangle SRQ$ ,

$TP=SR$ [Given]

$TQ=SQ$ [Given]

$PQ=QR$ [Given]

Therefore, by the

$SSS$ rule of congruence,

$\triangle TPQ\cong \triangle SRQ$

So, pairs of corresponding angles are,

$\angle T=\angle S$

$\angle TPQ=\angle SRQ$

$\angle TQP=\angle SQR$ .

Decide whether the SSS congruence is true with the following figures. Give reasons

From given figure,

$AC=EB, CR=BN, AR=NE$

Therefore, SSS congruence is true for Triangle

$ACR$ and Triangle

$EBN$

Complete the congruence statement.
$\Delta QRS \cong$

Solution:

In the given figure,

$QR=TP$

$QS=TQ$

$RS=PQ$

$\angle P=\angle R$

$\implies \triangle QRS \cong \triangle TPQ$ (by SSS congruence)

$\therefore$ we get the following correspondence in the vertices

$Q\leftrightarrow T$

$R\leftrightarrow P$

$S\leftrightarrow Q$

Hence,

$\bigtriangleup QRS \cong \bigtriangleup TPQ$

You want to show that $\Delta ART \cong \Delta PEN$ ,
If it is given that $\angle T = \angle N$ and you are to use SAS criterion, you need to have
(a) RT =
(b) PN =

Given that

$\Delta ART\cong \Delta PEN$ and

$\angle T=\angle N$

If the SAS (side angle side) criterion is used, then corresponding sides which are adjacent the equivalent angle of the triangle are equivalent.

Therefore,

$RT=EN$ and

$PN=AT$

In Fig. $AC = AE, AB = AD$ and $\angle BAD = \angle EAC$ . Show that $BC = DE$ .

In the given figure,

$AC=AE$

$AB=AD$

$\angle BAD=\angle EAC$

To prove:- $BC=DE$

Proof:-

Given that,

$\angle BAD=\angle EAC$

Adding both side $\angle DAC$ and we get,

$\angle BAD+\angle DAC=\angle EAC+\angle DAC$

$\angle BAC=\angle DAE\,\,.......\,\left( 1 \right)$

Now, In $\Delta ABC$ and $\Delta ADE$ ,

$AC=AE\,\,\,\,\,\,\,\,\,\,\left( Given \right)$

$\angle BAC=\angle DAE\,\left( by\,equation\,1 \right)$

$AB=AD\,\,\,\,\,\,\left( Given \right)$

Then, $\Delta ABC\cong \Delta ADE$ (S.A.S. Congruency)

$\therefore \,BC=DE$

Hence proved.

In the given figure, if $AC=DC$ and $CB=CE$ , then prove that $AB=DE$ (State Euclid's A Used

Given,

$AC=DC$ and

$CB=CE$

Consider, the line

$AB$

$AB=AC+CB$

$AB=DC+CE$ ------- from the given data

$\therefore AB=DE$

$O$ is a point in the interior of a square $ABCD$ such that $OAB$ is an equilateral triangle show that $\triangle OCD$ is an isosceles triangle

Given,

$\Delta OAB$ is an equilateral triangle,

$\angle AOB=\angle OAB=\angle OBA=60^\circ$

$OA=AB=OB$

Also it's given,

$ABCD$ is a Square

$\angle DCB=\angle BAD=\angle ABD=\angle CDA=90^\circ$ ------ (ii)

$DA=AB=CB=CD$

Consider,

$\angle OAB=\angle OBA=60^\circ$ ------ (i)

$\angle BAD=\angle ABD=90^\circ$ ------ (ii)

Subtracting (i) from (ii), we get,

$\angle BAD-\angle OAB=\angle ABD-\angle OBA=90^\circ-60^\circ=30^\circ$

i.e.,

$\angle DAO=\angle CBO=30^\circ$ ------ (iii)

Now, in

$\triangle AOD$ and

$\triangle BOC$

$AO=BO$ [

$\Delta OAB$ is an equilateral triangle]

$\angle DAO=\angle CBO=30^\circ$ [from (iii)]

$AD=BC$ [

$ABCD$ is a square]

Hence, by

$SA$ congruence criterion,

$\triangle AOD \cong \triangle BOC$

Hence,

$DO=OC$

$\therefore \Delta OCD$ is an isosceles triangle.

$\quad \quad [Hence\ proved]$

1148425_1024824_ans_70ce9d8f403a48cca3988c4613d97513.png

Prove that the lines:
$(a+b)x+(a-b)y-2ab=0.......(1)$
$(a-b)x+(a+b)y-2ab=0.......(2)$
and $x+y=0......(3)$ form an isosceles triangle whose vertical angle is $2\tan ^{ -1 }{ \left( a/b \right) }$ .

Slopes of the given lines are respectively

$-\cfrac { a+b }{ a-b } ,-\cfrac { a-b }{ a+b } ,-1$
The triangle will be isosceles if

$x+y=0$ is equally inclined to other two whose slopes are written above

$\cfrac { -1+\left( \cfrac { a+b }{ a-b } \right) }{ 1+\left( \cfrac { a+b }{ a-b } \right) } =-\cfrac { -1+\left( \cfrac { a+b }{ a-b } \right) }{ 1+\cfrac { a-b }{ a+b } } =\tan { \theta } \quad$

$\therefore \tan { \theta } =b/a\quad$

$\therefore$

$\triangle$ is isosceles whose vertical angle is

$\pi -2\theta =2\left( \cfrac { \pi }{ 2 } -\tan ^{ -1 }{ \cfrac { b }{ a } } \right) =2\left( \cot ^{ -1 }{ \cfrac { b }{ a } } \right) =2\tan ^{ -1 }{ \cfrac { a }{ b } }$
1029290_1007282_ans_01e9cd15b8d447838fc88db2630cad52.PNG

1029290_1007282_ans_01e9cd15b8d447838fc88db2630cad52.PNG

In a right angled triangle, prove that the line segment joining the mid-point of the hypotenuse to the opposite vertex is half the hypotenuse.

Let

$P$ be the mid point of the hypotenuse of the right

$\triangle ABC$ right angled at

$B$

Draw a line parallel to

$BC$ from

$P$ meeting

$B$ at

$O$

Join

$PB$

$\triangle PAD$ and

$\triangle PBD$

$\angle PDA=\angle PDB { 90 }^{ \circ }$ each due to conv of mid point theorem

$PD=PD$ ....... (common)

$AD=DB$ ...... (As

$D$ is mid point of

$AB$ )

$\triangle$

$PAD$ and

$PBD$ are congruent by SAS rule

$PA=PB$ ...... (C.P.C.T)

$PA=PC$ .... (Given as

$P$ is mid-point)

$\therefore PA=PC=PB$

Prove that in a $\triangle PQR$ if $QR^{2} = PQ^{2} + PR^{2}$ then $\angle P$ is a right angle.

Construction:

Draw a

$\Delta ABC$ right angled at

$A$ such that

$AB=PQ$

$CA=RP$

Now

$\Delta ABC$ ,

$\angle A=90^{\circ}$

By Pythagoras Theorem,

$(BC)^{2}=(CA)^{2}+(AB)^{2}$

$AB=PQ$ and

$CA=RP$

$(BC)^{2}=(RP)^{2}+(PQ)^{2}$ (1)

Given:

$(QR)^{2}=(RP)^{2}+(PQ)^{2}$ (2)

Hence from (1) and(2)

$BC=QR$

$\Delta ABC$ and

$\Delta PQR$

$AB=PQ$ (from construction)

$CA=RP$ (from construction)

$BC=QR$

$\Delta ABC\cong\Delta PQR$ (By SSS congruence)

$\angle A=\angle P=90^{\circ}$

Hence proved.

Prove that the inequality $R \geqslant 2r$ (R and r are the radii of the circumscribed and the inscribed circle) holds true in any triangle, and the equality R = 2r holds only. for a regular triangle.

1168634_890505_ans_bf182973f3aa436684e5045ecf8a1343.jpg

I and $m$ are two parallel line intersected by another pair of parallel line $p$ and $q$ Fig Show taht $\Delta ABC \cong \Delta CDA$ .

1968403_1036778_ans_e8facafd411741b5beec120ee992db5f.jpg

In the adjoining figure, $D$ and $E$ are points on the side $BC$ of $\triangle ABC$ such that $BD=EC$ and $AD=AE$ . Show that $\triangle ABD\cong \triangle ACE$ .

$AD=AE\Rightarrow \angle ADE=\angle AED$

$\Rightarrow {180}^{o}-\angle ADE={180}^{o}-\angle AED$

$\Rightarrow \angle ADB=\angle AEC$ .

$\triangle ABD\cong \triangle ACE$ (SAS).

A round balloon of radius r subtends an angle $\alpha$ at the eye of the observer, while the angle of elevation of its centre is $\beta$ . Prove that the height of the centre of the balloon is $r \sin\beta\ \text{csc}\dfrac{\alpha}{2}$ .

Let

$O$ be the centre of the balloon and

$P$ be the eye of the observer

And

$\angle APB$ be the angle subtended by the balloon

$\therefore \angle APB=\alpha$

$\therefore \angle APO=\angle BPO=\cfrac { \alpha }{ 2 }$

$\triangle OAP,\sin { \cfrac { \alpha }{ 2 } } =\cfrac { OA }{ OP }$

$\implies \sin { \cfrac { \alpha }{ 2 } } =\cfrac { r }{ OP }$

$\implies OP=r\text{ cosec }{ \cfrac { \alpha }{ 2 } } -(1)$

$\triangle OPC$

$\sin { \beta } =\cfrac { OL }{ OP }$ or

$OL=OP\sin { \beta }$

$\therefore OL=r\text{ cosec }{ \cfrac { \alpha }{ 2 } } \sin { \beta }$

1018069_1028152_ans_47ef5ef3db57453eb4c03e54e196ddf0.png

Show by diagram any two real-life examples for congruent shapes.

1. Pair of ear-rings

2. Pages of same book

If O is a point within a quadrilateral ABCD, prove that ; $OA + OB + OC + OD > AC + BD$ .

$\Delta OAC$

$OA+OC > AC$ _____ (1)

(sum of two sides of a

$\Delta$ is greater than third side)

similarity in

$\Delta OBD$

$OB+OD > BD$ _______ (2)

(1)+(2)

$OA+OB+OC+OD > AC+BD$

1048987_1053292_ans_8422e164d4ba411dbcb273003b265869.png

It is to be established by $RHS$ congruence rule that $\triangle ABC\cong \triangle RPQ$ . What additional information is needed, if it is given that $\angle B=\angle P=90^o$ and $AB=RP$ ?

For two right angle triangles to be congruent, we need to show two corresponding parameters as equal, with one of them as side for sure.

$\begin{array}{l} { { Since } },\, \, \Delta ABC\cong \Delta RPQ \\ and\; AB=RP \\ Any\ one\ of\ these\ conditions \ is\ required:\\ \;\ AC=RQ\\ \; BC=PQ \\ \angle BAC=\angle PRQ \\ and\, \, \angle ACB=\angle RQP \end{array}$

1218935_1053160_ans_7edabefe947c4c2da631a2af41985e87.png

1218935_1053160_ans_7edabefe947c4c2da631a2af41985e87.png

In the following figure $\angle B < \angle A$ and $\angle C < \angle D$ , show that $AD < BC$ .

From figure.

$AO < BO$ (side opposite to greater angle is long )

$OD < OC (\quad " \quad " \quad )$

$AO + OD < BO + OC$

$AD < BC$ Hence proved.

1063138_1038315_ans_59f3a45c79bf478d91131e12ecdfd833.jpg

Let $x,\ x+1,\ x+2$ , be the lengths of the three sides of a triangle.
(i) Write down the three inequations in x, each of which represents the given statement.
(ii) List the set of possible values of x which satisfy all the three inequations obtained in your answer to part (i) above, given that x is an integer.

$(i)x+(x+1)>(x+2)$

$=>2x+1>x+2$

$=>x>1 -(1)$

$(x+1)+(x+2)>x$

$=>2x+3>x$

$=>x>-3 -(2)$

$x+(x+2)>x+1$

$=>2x+2>x+1$

$=>x>-1 -(3)$

$(1)(2)\& (3) represent the three inequalities$

$(ii)comparing (1),(2)\& (3)$

$=> x>-3$

$x>-1$

$x>1$

$\therefore x>1$

$\therefore possible values of x=2,3,4,5,6,......$

Given that $ABCD$ is a quadrilateral. Is $AB+BC+CD+DA<2(AC+BD)?$

As we know that the sum of lengths of any two sides in a triangle should be greater than the length of third side-

Therefore,

In $\triangle AMB, \ AB < MA + MB$ …….(i)

In $\triangle BMC, \ BC < MB + MC$ …….(ii)

In $\triangle CMD,\ CD<MD+MC$ ..….(iii)

In $\triangle AMD,\ DA < MD + MA$ …….(iv)

On adding equation (i), (ii), (iii) and (iv), we get

$AB + BC + CD + DA < 2MA + 2MB + 2MC + 2MD$

$AB + BC + CD + DA < 2[(AM + MC) + (DM + MB)]$

Since, $AC=AM+MC\ and\ BD=BM+MD$

Hence,

$AB + BC + CD + DA < 2(AC + BD)$

It is proved.

987520_1045943_ans_87a54a57f44247998601dd7a7433327a.jpg

If $AB>AC$ and length of median from $C$ is $4$ units, then the length of median from $B$ can be-

$CN=\cfrac { 1 }{ 2 } \sqrt { 2({ AC }^{ 2 }+{ BC }^{ 2 })-{ AB }^{ 2 } }$

$\implies\quad Y=\cfrac { 1 }{ 2 } \sqrt { 2({ AC }^{ 2 }+{ BC }^{ 2 })-{ AB }^{ 2 } }$

$\implies\quad 2({ AC }^{ 2 }+{ BC }^{ 2 })-{ AB }^{ 2 }=64\quad$ -(1)

$\implies\quad 2{ BC }^{ 2 }=64+{ AB }^{ 2 }-{ 2AC }^{ 2 }\quad$ -(2)

$BP=\cfrac { 1 }{ 2 } \sqrt { 2({ AB }^{ 2 }+{ BC }^{ 2 })-{ AC }^{ 2 } }$

$\implies\quad BP=\cfrac { 1 }{ 2 } \sqrt { 2({ AB }^{ 2 }+{ BC }^{ 2 })-{ AC }^{ 2 } }$

using(2)

$\implies\quad BP=\cfrac { 1 }{ 2 } \sqrt { 2{ AB }^{ 2 }+64+{ AB }^{ 2 }-{ 3AC }^{ 2 } }$

$\implies\quad BP=\cfrac { 1 }{ 2 } \sqrt { 64+3({ AB }^{ 2 }-{ AC }^{ 2 }) }$

$\because$ AB>AC

$\therefore$ AB-AC>0 -(3)

$\implies\quad BP=\cfrac { 1 }{ 2 } \sqrt { 64+3(AB-AC)(AB+AC) }$ using(3)

$BP=\cfrac { 1 }{ 2 } \sqrt { 64+3(AB-AC)(AB+AC) } >\cfrac { 1 }{ 2 } \sqrt { 64+0 }$

$\implies BP>\cfrac { 1 }{ 2 } \times 8$

$\implies$ BP>4 (BP=length of median from B)

1004679_1044522_ans_3d2247cb20804872ade6a89568f2addd.png

In the given figure, $\triangle CDE$ is an equilateral triangle on a side $CD$ of a square $ABCD$ . Show that $\triangle ADE \cong \triangle BCE$ .

According to figure,

$\begin{array}{l}\angle ADC = \angle BCD = {90^ \circ } \to square \\\angle CDE = \angle DCE = {60^ \circ } \to equilateral\_triangle\\\therefore \angle ADE = \angle BCE = {150^ \circ }\\In \Delta ADE and \Delta BCE \\AD = BC\\DE = CE\\\angle ADE = \angle BCE\\\Delta ADE \cong \Delta BCE\end{array}$

$AB$ is a line segment and $P$ is its mid-point. $D$ and $E$ are points on the same side of $AB$ such that $\angle BAD=\angle ABE$ and $\angle EPA=\angle DPB$ . Show that
i) $\triangle DAP\cong \triangle EBP$
ii) $AD=BE$

As shown in the figure,

$\angle EPA=\angle DPB$

So,

$\angle APE+\angle EPD=\angle BPD+\angle DPE$

Hence,

$\angle APD=\angle BPE$ ....(1)

Opposite side of

$\angle APD$ is

$AD$ and opposite side of

$\angle DPE$ is

$BE$ ,

From equation (1),

$AD=BE$ ......(2)

Hence proved.

$AP=PB$ ......(3)

$\triangle DAP\ and\ \triangle EBP$ ,

$\angle APD=\angle BPE$ ....(1)

$AD=BE$ ......(2)

$AP=PB$ ......(3)

Hence,

$\triangle DAP\cong \triangle EBP$ by (S.A.S). congurency

It's proved.

Given :- $AC=AD$
$\angle CAB= \angle BAD$
To prove :- $\triangle ABC \cong \triangle ABD$
Proof :- $9n \triangle ABC$ & $\triangle ABD$
$AC=AD$
$\angle CAB= \angle BAD$
$AB=BA$
$\triangle ABC \cong \triangle ABD$
$BC=BD$

$\begin{array}{l} Given\, \, \angle CAB=\angle BAD \\ In\, \, \Delta CAB\, \, and\, \, DAB \\ AB=AB\, \, \left( { common } \right) \\ \angle CAB=\angle DAB\, \, \left( { given } \right) \\ AC=AD\, \, \left( { given } \right) \\ By\, \, S.A.S.\, \, congruency \\ \Delta CAB\cong DAB \\ \therefore BC=BD \end{array}$
1223235_1052261_ans_4161ec8f3cbb4fb9a6b7b0733a330797.JPG

1223235_1052261_ans_4161ec8f3cbb4fb9a6b7b0733a330797.JPG

Prove that sum of any two sides of a triangle is greater than the third side.

Construction: In $\Delta ABC$ , extend $AB$ to D in such a way that $AD = AC$ .

In $\Delta DBC$ , as the angles opposite to equal sides are always equal, so,

$\angle ADC = \angle ACD$

Therefore,

$\angle BCD > \angle BDC$

As the sides opposite to the greater angle is longer, so,

$BD > BC$

$AB + AD > BC$

Since $AD = AC$ , then,

$AB + AC > BC$

Hence, sum of two sides of a triangle is always greater than the third side.

991253_1057262_ans_b55b2b265e4a4a0dbc07d8544c21f39b.PNG

In the adjoining figure. AB = AC and BD = DC. Prove that $\Delta ADB\, \cong \,\Delta ADC$ and hence show that

(i) $\angle ADB\, = \,\angle ADC\,\,$

(ii) $\angle BAD\, = \,\angle CAD$

Given

$AB=AC,BD=DC$

$\triangle ADB,\triangle ADC$

$AB=AC$

$BD=DC$

$AD=AD$

$SSS$ congruency

$\triangle ADB\cong \triangle ADC$

$\implies \angle ADB=\angle ADC$

and also

$\angle BAD=\angle CAD$

The length of two sides of a triangle are $5\ cm$ and $7\ cm$ . Between what two measures should the length of the third side fall?

It is known that the sum of the two sides of a triangle is greater than the third side, so,

${\rm{third}}\;{\rm{side}} < 5 + 7$

${\rm{third}}\;{\rm{side}} < 12$

And,

${\rm{third}}\;{\rm{side}} > 7 - 5$

${\rm{third}}\;{\rm{side}} > 2$

Therefore, $2 < {\rm{third}}\;{\rm{side}} < 12$ .

Hence, the third side of the triangle falls between $\left( {2,12} \right)$ .

In the example given, a pair of triangles is shown. Equal parts of triangles in pairs are marked with the same signs. Observe the figures and state the test by which the triangles in pair are congruent.

$AB = PQ$

$BC = QR$

$AC = PR$

$SQ$ by SSS property

$\Delta ABC \cong \Delta PQR$

1333777_1063756_ans_9eda3f61d4a04a3ca8b1fb0645e97a42.png

In triangles $ABC$ and $PQR, \angle A = \angle Q$ and $\angle B = \angle R$ . Which side of $\Delta PQR$ should be equal to side $BC$ of $\Delta ABC$ so that the two triangle are congruent? Give reason for your answer.

First draw the triangles and given

$\angle A=\angle Q$ &

$\angle B=\angle R$

So, to prove that two triangles are congruent.

$BC=RP$

So, we can say that

$\Delta ABC\cong \Delta QRP$

So, we can say that

$RP$ of

$\Delta PQR$ should be equal to side

$BC$ of

$\Delta ABC$ .

1259807_1055841_ans_82045e91fce44851babdf5614888c54b.png

"If two angles and a side of one triangle are equal to two angles and a side of another triangle, then the two triangles must be congruent". Is the statement true? Why?

The statement is true because, the triangles will be congruent either by ASA rule or AAS rule. This is because two angles and one side are sufficient to construct two congruent triangles.

Line $l$ is the bisector of an angle $\angle{A}$ and $B$ is any point on $l.\,\,BP$ and $BQ$ are perpendiculars from $B$ to the arms of $\angle{A}$
Show that
$(i)\triangle{APB}\cong\triangle{AQB}$
$(ii)BP=BQ$ or $B$ is equidistant from the arms of $\angle{A}$

Given:

$l$ is the bisector of

$\angle{A}$

So,

$\angle{PAB}=\angle{QAB}$ .......

$(1)$

$BP$ and

$BQ$ are perpendiculars from

$B,$

So,

$\angle{APB}=\angle{AQB}={90}^{\circ}$ ........

$(2)$

$\triangle{APB}$ and

$\triangle{AQB},$

$\angle{APB}=\angle{AQB}$ from

$(2)$

$\angle{PAB}=\angle{QAB}$ from

$(1)$

$AB=AB$ (common)

$\therefore \triangle{APB}\cong\triangle{AQB}$ [by

$AAS$ congruency rule]

$\therefore BP=BQ$ [by

$CPCT$ ]

Hence proved.

S is any point on side QR of a $\Delta PQR$ . Show that : PQ + QR + RP > 2 PS

Theorem: In a triangle , sum of the length of any two sides is greater than the third side.

Draw the triangle and join the points

$P$ and

$S$

In the figure,

$\Delta PQS$ , according to the theorem,

$PQ+QS>PS$ ........... (1)

$\Delta PSR$ , according to the theorem,

$PR+SR>PS$ ........... (2)

Adding (1) and (2),

$PQ+QS+SR+PR>PS+PS$

$PQ+(QS+SR)+PR>2PS$

$PQ+QR+PR>2PS$

980644_1069768_ans_96846ff94e4c47df9c71680bf4ad99a5.png

In $\Delta$ ABC, side $AB$ is produced to $D$ such that $BD=BC$ . If $\angle CAB=70^o$ and $\angle CBA=60^o$ , prove that $AD>AC$ .

Given,

$\angle CAB={ 70 }^{ \circ },\angle CBA={ 60 }^{ \circ }$

So,

$\angle CBD={ 180 }^{ \circ }-\angle CBA$

$={ 120 }^{ \circ }$

$\angle CBA$ is an exterior angle of

$\triangle BCD$

$\triangle BCD$ ,

$BC=BD$ , so it is an isosceles triangle.

$\angle BCD=\angle BDC=\dfrac{ \angle CBA}{2}={ 30 }^{ \circ }$

Now, In

$\triangle ADC$

$\angle A={ 70 }^{ \circ },\angle D={ 30 }^{ \circ }$

So,

$\angle C={ 180 }^{ \circ }-({ 70 }^{ \circ }+{ 30 }^{ \circ })={ 80 }^{ \circ }$

Side

$AD$ is opposite to

$\angle C$ and side

$AC$ is opposite to

$\angle D$

$\because \angle D<\angle C\\ \Rightarrow AC<AD\quad or\quad AD>AC$

[

$\because$ side opposite to greater

$\angle$ is longer]

Hence,

$proved$ .

$AB$ is a line segment and $P$ is its mid-point. $D$ and $E$ are points on the same side of $AB$ such that $\angle BAD =\angle ABE$ and $\angle EPA =\angle DPB$ . Show that $\Delta DAP \cong \Delta EBP$ .

$\angle BAD=\angle ABE\Rightarrow \angle PAD=\angle PBE$

$\angle EPA=\angle DPB$

$\Rightarrow \angle DPA+\angle EPD=\angle EPD+\angle EPB$

$\Rightarrow \angle DPA=\angle EPB$

$\therefore$ In

$\Delta DAP$ and

$\Delta EBP$

$\angle PAD=\angle PBE - (i)$

$AP=BP - (ii)$ (given)

$\angle DPA=\angle EPB - (iii)$

$\therefore \Delta DAP \cong \Delta EBP$ by ASA test of congruence.

1075132_1072439_ans_b957ae8c531249c187740eabd8127995.png

In $\Delta$ ABC, side AB is produced to D such that BD $=$ BC. If $\angle A=70^o$ and $\angle B=60^o$ , prove that AD $>$ CD.

$\angle ACB={ 50 }^{ \circ }$ Sum of angles

$= 180^\circ$

Let

$\angle BCD=x\quad ,\quad \angle BDC=y$

now,

$x+y={ 60 }^{ \circ }$ [exterior angle property]

Also,

$70+50+x+y={ 180 }^{ \circ }$ Sum of angles

$= 180^\circ$ for triangle ACD

Given

$BD=BC$

$\Rightarrow x=y={ 30 }^{ \circ }$ Two sides of same length hence same angle.

$\angle ACD={ 50 }^{ \circ }+{ 30 }^{ \circ }={ 80 }^{ \circ }$

Now, in

${ \Delta}ACD, CD$ is the side opposite to

$\angle A$ &

$AD$ is side opposite to

$\angle C$ .

$\because \angle C>\angle A$

$\therefore AD>CD\quad [\because$ side opposite to greater angle is longer

$]$

$AB$ is a line segment and $P$ is its mid-point. $D$ and $E$ are points on the same side of $AB$ such that $\angle BAD =\angle ABE$ and $\angle EPA =\angle DPB$ . Show that $AD=BE$ .

$\angle BAD=\angle ABE\Rightarrow \angle PAD=\angle PBE$

$\angle EPA=\angle DPB$

$\Rightarrow \angle DPA+\angle EPD=\angle EPD+\angle EPB$

$\Rightarrow \angle DPA=\angle EPB$

$\therefore$ In

$\Delta DAP$ and

$\Delta EBP$

$\angle PAD=\angle PBE - (i)$

$AP=BP - (ii)$

$\angle DPA=\angle EPB - (iii)$

$\therefore \Delta DAP \cong \Delta EBP$ by ASA test of congruence

$\because$ Two triangle are congruent

$\therefore$ Corresponding sides should be equal

$AD = BE$ (proved)

1075127_1072441_ans_a12c3373f35a4e10850fe11332d5c73a.png

In the figure $AB=AC,BD=DC,\angle C=40^{o}\angle BDC=160^{o}$
Write the measure of all angles of triangle $ABD$ .

Consider the following question.

Given that.

$AB=AC,BD=DC,\angle C={{40}^{0}}\angle BDC={{160}^{0}}$

Find the value of $\Delta ABD=?$

We know that,

$\angle BAC=\dfrac{1}{2}\angle BDC$

$=\dfrac{1}{2}\times {{160}^{0}}$

$\angle BAC={{80}^{0}}$

In $\Delta CAD$ ,

Therefore,

$\angle ADC+\angle DCA+\angle CAD={{180}^{0}}$

$\angle ADC+{{40}^{0}}+{{40}^{0}}={{180}^{0}}$

$\angle ADC={{100}^{0}}$

In $\Delta ABD$

$\angle BAD=360-(\angle BDC+\angle CDA)$

$={{360}^{0}}-\left( {{160}^{0}}+{{100}^{0}} \right)$

$\angle BAD={{100}^{0}}$

$\angle DAB={{40}^{0}}$

$\angle BAD+\angle DAB+\angle ABD={{180}^{0}}$

${{40}^{0}}+{{100}^{0}}+\angle ABD={{180}^{0}}$

$\angle ABD={{40}^{0}}$

Hence, this is the required answer.

$D$ is any point on side $AC$ of a $\Delta ABC$ with $AB = AC$ . Show that $CD < BD.$

Consider the diagram shown below.

Given:

$AB=AC$

To prove:

$CD<BD$

Proof:

$\angle ABC=\angle ACB$ (Isosceles Triangle) …… (1)

Here,

$\angle ABC = \angle CBD+\angle DBA$

$\therefore \angle ABC>\angle CBD$

Also, $\Rightarrow \angle ACB = \angle DCB>\angle CBD$ {Because $\angle ABC = \angle DCB = \angle ACB$ }

Therefore,

$CD<BD$ (Converse of Angle opposite to greater side is greater in a triangle)

Hence, proved.

1037414_1071247_ans_0c6f5093e23d446886f399daded5f64c.png

$ABCD$ is a quadrilateral in which $AD=BC$ and $\angle DAB=\angle CBA.$ Prove that $BD =AC.$

$\triangle ABD$ and

$\triangle BAC,$

$AD=BC$ [ Given ]

$\angle DAB=\angle CBA$ [ Given ]

$AB=BA$ [ Common side ]

So, by the Side-Angle-Side criterion of congruence,

$\triangle ABD\cong \triangle BAC$

$\therefore$

$BD=AC$ [CPCT]

Let $ABC$ be an isosceles triangle with $AB=AC$ and let $D, E, F$ be the mid-point of $BC, CA$ and $AB$ respectively. Show that $AD\bot EF$ and $AD$ is bisected by $EF$ .

1326238_1086496_ans_c88c0edf5d3b4f44bfcc46a8025e720d.jpg

AD is an altitude of an isosceles triangle ABC in which $AB=AC$ . show that
(i) AD bisects BC
(ii) AD bisects $\angle A$

$\Delta ABC$

$AD\perp BC$

$AB=AC........$ (Given)

$AD=AD.......$ (Common Side)

$\angle ADB=\angle ADC=90^{\circ}$

$RHS$ criteria

$\Delta ADB\cong \Delta ADC$

$\implies BD=DC\implies AD$ bisects

$BC$

$\implies \angle BAD=\angle CAD\implies AD$ bisects

$\angle A$

1318850_1067389_ans_e7619cb29f90421eaef9d54b5c3e8ce1.png

AM is a median of a triangle $ABC$ .
Is $AB + BC + CA > 2 AM$ ?
(Consider the sides of triangles $\Delta ABM$ and $\Delta ABM$ .)

Triangle ABM

$AB+BM > AM$ ……….

$(1)$ (Sum of

$2$ sides

$>$

$3^{rd}$ side)

In triangle ACM

$AC+CM > AM$ ……….

$(2)$

By adding

$1$ &

$2$

$(AB+BM)+(CM+AC) > 2AM$

$AB+AC+(BM+CM) > 2AM$

$AB+AC+BC > 2 AM$

$\therefore$ Hence proved.

In the given figure, $\angle B < \angle A$ and $\angle C < \angle D$ , show that $AD < BC$ .

Let the point of intersection of

$BC$ and

$AD$ be

$O$

$\triangle ABO$

$\Rightarrow \angle B<\angle A$

Thereore,

$\Rightarrow AO<BO$

$(1)$ (side opposite to greater angle is largest)

Similarly,

$\triangle COD$

$\Rightarrow \angle C<\angle D$

Thereore,

$\Rightarrow DO<CO$

$(2)$ (side opposite to greater angle is largest)

Adding

$(1)$ and

$(2)$ we get,

$\Rightarrow BO+OC>AO+DO$

$\Rightarrow BC>AD$

Hence proved.

998312_1091066_ans_99c6670e75f64266ac8115ab41633308.png

In Fig., $D$ and $E$ are points on side $BC$ of a $\triangle ABC$ such that $BD = CE$ and $AD = AE$ . Show that $\triangle ABD \cong \triangle ACE$ .

1156620_1124098_ans_973efd605ec344968b4acd275689bd08.jpg

Prove that any two sides of a $\Delta$ are together greater than twice the median drawn to the third side.

Given:Triangle ABC in which AD is the median.

To prove:

$AB+AC>2AD$

Proof:In

$ΔADB$ and

$ΔEDC$

$AD=DE$ [By construction]

D is the midpoint BC.[$$DB=DB$$]

$ΔADB=ΔEDC$ [vertically opposite angles]

Therefore

$ΔADB≅ ΔEDC$ [By SAS congruence criterion.]

$AB=ED$ [Corresponding parts of congruent triangles]

$ΔAEC$ ,

$AC+ED>AE$ [sum of any two sides of a triangle is greater than the third side]

$AC+AB>2AD$ [

$AE=AD+DE=AD+AD=2AD$ and

$ED=AB$ ]

1121479_1111803_ans_244c6540ee3f4efaadbb482f51b4d72d.png

In quadrilateral $ACBD$
$AQC=AD$ and $AB$ bisects $\angle{A}$ (see figure). Show that $\triangle{ABC}\cong \triangle{ABD}$ .
What can you say about $BC$ and $BD$ ?

$\Delta ACB$ and

$\Delta ADB$

$side \, AB = side \, AB$ ......common side

$\angle CAB = \angle DAB$ .....(angle bisector)

side

$CA=$ side

$DA$ .....given

$\therefore \Delta ACB \equiv \Delta ADB$ by

SAS test

also

$BC=BD$ .....(since both triangles are congruent)

In figure, $ABC$ is a triangle with $\angle B = 35^{\circ}, \angle C = 65^{\circ}$ and bisector of $\angle BAC$ meets $BC$ in $X$ . Arrange $AX, BX$ and $CX$ in descending order.

Consider

$\triangle ABC$

By sum property of a triangle

$\angle A + \angle B + \angle C = 180^{\circ}$

To find

$\angle A$

$\angle A = 180^{\circ} - \angle B - \angle C$

By substituting the values

$\angle A = 180^{\circ} - 35^{\circ} - 65^{\circ}$

By subtraction

$\angle A = 180^{\circ} - 100^{\circ}$

$\angle A = 80^{\circ}$

We know that

$\angle BAX = \dfrac {1}{2} \angle A$

So we get

$\angle BAX = \dfrac {1}{2} (80^{\circ})$

By division

$\angle BAX = 40^{\circ}$

Consider

$\triangle ABX$

It is given that

$\angle B = 35^{\circ}$ and

$\angle BAX = 40^{\circ}$

By sum property of a triangle

$\angle BAX + \angle BXA + \angle XBA = 180^{\circ}$

To find

$\angle BXA$

$\angle BXA = 180^{\circ} - \angle BAX - \angle XBA$

By substituting values

$\angle BXA = 180^{\circ} - 35^{\circ} - 40^{\circ}$

By subtraction

$\angle BXA = 180^{\circ} - 75^{\circ}$

$\angle BXA = 105^{\circ}$

We know that

$\angle B$ is the smallest angle and the side opposite to it i.e.

$AX$ is the smallest side.

So we get

$AX < BX .. (1)$

Consider

$\triangle AXC$

$\angle CAX = \dfrac {1}{2} \angle A$

So we get

$\angle CAX = \dfrac {1}{2} (80^{\circ})$

By division

$\angle CAX = 40^{\circ}$

By sum property of a triangle

$\angle AXC + \angle CAX + \angle CXA = 180^{\circ}$

To find

$\angle AXC$

$\angle AXC = 180^{\circ} - \angle CAX - \angle CXA$

By substituting values

$\angle AXC = 180^{\circ} - 40^{\circ} - 65^{\circ}$

So we get

$\angle AXC = 180^{\circ} - 105^{\circ}$

By subtraction

$\angle AXC = 75^{\circ}$

So we know that

$\angle CAX$ is the smallest angle and the side opposite to it i.e.

$CX$ is the smallest side.

We get

$CX < AX (2)$

By considering equation

$(1)$ and

$(2)$

$BX > AX > CX$

Therefore,

$BX > AX > CX$ is the descending order.

$ABCDE$ is a regular pentagon having all the sides and angles equal. Show that $\triangle ABC \cong \triangle AED$ .

Given:

$BCDEA$ is a regular Pentgon. All sides and angles are equal

Therefore,

$\triangle ABC$ and

$\triangle AED$ ,

$1)\ AB=AE$

$2) BC=ED$

$3) \angle ABC= \angle AED$

$\therefore \ \triangle ABC \cong \triangle AED$ (

$SAS$ congruency)

Hence proved.

$ABC$ is an isosceles triangle with $AB = AC$ and $BD$ and $CE$ are its two medians. Show that $BD = CE$ .

Given:-

AB = AC

Also , BD and CE are two medians

Hence ,

E is the midpoint of AB and

D is the midpoint of CE

Hence ,

1/2 AB = 1/2AC

BE = CD

In Δ BEC and ΔCDB ,

BE = CD [ Given ]

∠EBC = ∠DCB [ Angles opposite to equal sides AB and AC ]

BC = CB [ Common ]

Hence ,

Δ BEC ≅ ΔCDB [ SAS ]

BD = CE (by CPCT)

In the square $ABCD$ show that the two triangles $ABC$ and $ADC$ are congruent to each other

we have,

In the square $ABCD$

Construct:- Draw diagonal $AC$ then makes two $\Delta ABC\,\And \,\Delta ADC$

Prove that:- $\Delta ABC\cong \Delta ADC$

Proof:- In $\Delta ABC\,\And \,\Delta ADC$

$AB=DC\,\,\,\,\,\,\,\left( sides\,of\,square \right)$

$AD=BC\,\,\,\,\,\,\,\left( sides\,of\,square \right)$

$AC=AC\,\,\,\,\,\,\,\left( common \right)$

Hence, $\Delta ABC\cong \Delta ADC$ $\left( by\,SSS\,congruency \right)$

Hence proved.

1034946_1100098_ans_f8e13e162c2245878addd6871f4dca19.jpg

If two angles of a triangle are equal, then prove that sides opposite to them are also equal.

$\angle C=\angle B$

$\triangle ABD$ &

$\triangle ACB$

$\angle B=\angle C$ (Given)

$\angle BAD=\angle CAD$ (construction)

$AD=AD$

$\triangle ABD\cong \triangle ACD$

$AB=AC$

1378776_1147287_ans_2515c0a68f504837b1adebbff1d12464.png

In a triangle $ABC$ median $AD$ is produced to $X$ such that $AD+DX$ . Prove that $ABXC$ is a parallelogram.

Given that

$AD= DX$

Also

$BD= DC$ (because of median bisector that side of

$\triangle$ )

This show that diagonal

$AX$ and

$BC$ of

$\triangle AB \times C$ bisect each other at

$D$ which is property of

$11\ gram$

Thus

$AB \times C$ is a parallelogram

State whether the following triangles are congruent or not? Give reasons for your answer.

$\triangle FED$ ,

$\angle F+\angle E+\angle D={ 180 }^{ \circ }$ (Angles Sum property of a triangle)

${ 70 }^{ \circ }+\angle E+{ 60 }^{ \circ }={ 180 }^{ \circ }\\ \angle E={ 50 }^{ \circ }$

So, In

$\triangle ABC$ and

$\triangle DEF$ ,

Given that:

$\angle B= \angle E$

$BC=EF$

$\angle C= \angle E$

So,

$\triangle ABC\cong \triangle DEF$ {using

$ASA$ criterion}

$\triangle LMN$ and

$\triangle RTS$

Given that:

$\angle M= \angle T$

$MN=ST$

$NL=SR$

So,

$\triangle LMN\cong \triangle RTS$ {using

$ASS$ criterion}

In the adjoining figure, $BA \bot AC, \, DE \bot DF$ such that $BA = DE$ and $BF = EC$ . Show that $\Delta ABC \cong \Delta DEF$

$\triangle ABC,\triangle DEF$

$BA=ED$

$BF=EC$

$\implies BF+FC=FC+EC$

$BC=EF$

$\angle BAC=\angle EDF=90^{\circ}$

$S.A.S$ congruency

$\triangle ABC \cong \triangle DEF$

$ABCD$ is a quadrilateral
Prove that $(AB+BC+CD+DA)>(AC+BD)$

$(AB+BC+CD+DA)>(AC+BD)$

$ABCD$ is a quadrilateral and

$AC$ and

$BD$ are diagonals.

Sum of the two sides of

$\triangle$ is greater than the third side

$\triangle ABC, BCD, CAD$ and

$BAD$

$AB+BC>AC$

$CD+AD>AC$

$AB+AD>BD$

$BC+CD>BD$

Adding all the above equation

$2(AB+BC+CA+AD)>2(AC+BD)$

$\Rightarrow (AB+BC+CA+AD)>(AC+BD)$

1378052_1149045_ans_cedc547fd31549b38d5915f6d260865b.png

In $\Delta ABC,$ $AB<AC,$ $PB$ and $PC$ are the bisectors of $\angle B$ and $\angle C.$ Prove that $PB< PC.$

In the figure.

The angle opposite to a larger side is larger than the angle opposite to a smaller side.

It is given

$AB < AC$

$BP$ is the bisector of $\angle B$

So,

$\angle 1=\angle 2 =\dfrac12 \angle B$

$CP$ is bisector of $\angle C$

$\angle 3=\angle 4=\dfrac12 \angle C$

Angle opposite to $AB$ is $\angle C$ and angle opposite to $AC$ is $\angle B.$

Now, from the property

As $AB < AC$

So,

$\angle C<\angle B$

$\Rightarrow \dfrac12\angle C<\dfrac12 \angle B$

$\Rightarrow \angle 4<\angle 2$

Side opposite to $\angle 2$ is $PC$ and side opposite to $\angle 4$ is $PB.$

Hence,

$\angle 4<\angle 2$

$\Rightarrow PB < PC$

Hence, Proved.

1298464_1132187_ans_a563203d2dd5414ab1f6b9d3dcc72d71.jpg

$l, m$ and $n$ are three parallel lines intersected by transversals $p$ and $q$ such that $l, m$ and $n$ cut off equal intercepts $AB$ and $BC$ on $p$ . Show that $l, m$ and $n$ cut off equal intercepts $DE$ and $EF$ on $q$ also.

Given:

$l\parallel m\parallel n$

$l,m$ and

$n$ cut off equal intercepts

$AB$ and

$BC$ on

$p$

So,

$AB=BC$

To prove:

$l,m$ and

$n$ cut off equal intercepts

$DE$ and

$EF$ on

$q$

i.e.,

$DE=EF$

Proof:In

$\triangle{ACF},$

$B$ is the mid-point of

$AC$ as

$AB=BC$

and

$BG\parallel CF$ since

$m\parallel n$

So,

$G$ is the mid-point of

$AF$ using line drawn throught mid-point of one side of a triangle, parallel to another side, bisects the third side.

$\triangle{AFD},$

$G$ is the mid-point of

$AF$

and

$GE\parallel AD$ since

$l\parallel m$

So,

$E$ is the mid-point of

$DF$ using line drawn throught mid-point of one side of a triangle, parallel to another side, bisects the third side.

Since

$E$ is the mid-point of

$DF$

$DE=EF$

Hence proved.

If $\Delta$ MNO $\cong\Delta$ QPR, write the six corresponding parts.

$\begin{array}{l} \Delta MNO\cong \Delta QPR \\ MN=QP \\ MO=QR \\ NO=PR \\ and \\ \angle NMO=\angle PQR \\ \angle MNO=\angle QPR \\ \angle MON=\angle QRP \end{array}$
1213408_1137491_ans_20da041f6041402bb353905eb942d910.PNG

1213408_1137491_ans_20da041f6041402bb353905eb942d910.PNG

In the quadrilateral $ABCD,AD=CD$ and $\angle A ={90}^{o}=\angle C$ , prove that $AB=BC$ .

Reference to the image above,

$\triangle ABD$ &

$\triangle BCD$

$AD=CD$ (Given... height)

$BC= BC$ (Common... Hypotenuse)

$\angle BAD = \angle BCD$ (right angles)

$\therefore \triangle BAD \cong \triangle BCD$ (By RHS congruency Rule)

$\therefore AB= BC$ ( By CPCT)

$D$ is a point on the side $BC$ of a $\triangle ABC$ such that $\angle ADC=\angle BAC$ , then prove that $CA^2 = CB \times CD$

Data: D is a point on the side BC of a triangle

$\triangle ABC$ such that

$\angle ADC = \angle BAC.$

To Prove:

$CA^{2} = CB \times CD$
Let

$\angle ADC = \angle BAC = 100^\circ$
In

$\triangle ABC,$ if

$\angle B = 50^\circ,$ then

$\angle C = 30^\circ$
In

$\triangle ADC,$ if

$\angle C = 30^\circ,$ then

$\angle DAC = 50^\circ$
In

$\triangle BCP, \angle A = 100^\circ, \angle B = 50^\circ, \angle C = 30^\circ$
In

$\triangle ADC, \angle ADC = 100, \angle DAC = 50^\circ, \angle ACD = 30^\circ$
Similarity criterion of

$\triangle$ is A.A.A.

$\therefore$ In

$\triangle ABC$ and

$\triangle ADC,$

$\dfrac{CA}{BC} = \dfrac{DC}{CA}$

$\therefore CA \times CA = BC \times DC$

$\therefore CA^{2} = BC \times DC.$

1112088_1152327_ans_659cfc5b10b946f1b9ebc981a6e00219.png

The image of an object placed at a point A before a plane mirror $LM$ is seen at the point $B$ by an observer at $D$ as shown in Fig. Prove that the image is as far behind the mirror as the object is in front of the mirror.

According to the figure we need to prove that

$AT = BT$

We know that

Angle of incidence

$=$ Angle of reflection

So we get

$\angle ACN = \angle DCN .. (1)$

We know that

$AB \parallel CN$ and

$AC$ is the transversal

From the figure we know that

$\angle TAC$ and

$\angle ACN$ are alternate angles

$\angle TAC = \angle CAN (2)$

We know that

$AB \parallel CN$ and

$BD$ is the transversal

From the figure we know that

$\angle TBC$ and

$\angle DCN$ are corresponding angles

$\angle TBC = \angle DCN .. (3)$

By considering the equation

$(1), (2)$ and

$(3)$

We get

$\angle TAC = \angle TBC (4)$

Now in

$\triangle ACT$ and

$\triangle BCT$

$\angle ATC = \angle BTC = 90^{\circ}$

$CT$ is common i.e.

$CT = CT$

$AAS$ congruence criterion

$\triangle ACT \cong \triangle BCT$

$AT = BT (c. p. c. t)$

Therefore, it is proved that the image is as far behind the mirror as the object is in front of the mirror.

If $P$ is a point in the interior of $\triangle ABC$ then fill in the blanks with $>$ or $<$ or $=$ .
(i) $PA+PB.........AB$
(ii) $PB+PC........BC$
(iii) $AC.........PA+PC$

(i)

$PA+PB> AB$

Because, the sum of any two side of triangle is greater than the third side

(ii)

$PB+PC> BC$

Because, the sum of any two side of triangle is greater than the third side

(iii)

$AC< PA+PC$

Because, the sum of any two side of triangle is greater than the third side

Prove that the diagonals of a parallelogram bisect each other

${\textbf{Step - 1: Showing congruency}}$

${\text{In }}\vartriangle {\text{AOB and }}\vartriangle {\text{COD,}}$

$\angle {\text{1 = }}\angle {\text{2 (alternate angles)}}$

$\angle {\text{3 = }}\angle {\text{4 (alternate angles)}}$

${\text{AB = CD (opposite sides of a parallelogram are equal)}}$

${\text{By ASA, }}\vartriangle {\text{AOB }} \cong {\text{ }}\vartriangle {\text{COD}}$

${\textbf{Step - 2: Proving diagonals bisect each other}}$

${\text{Since }}\vartriangle {\text{AOB and }}\vartriangle {\text{COD are congruent,}}$

$\therefore {\text{ AO = OC and BO = OD}}$

$\Rightarrow {\text{ O is the midpoint of AC and BD}}$

$\Rightarrow {\text{ Diagonals of parallelogram bisect each other}}$

${\textbf{Hence, Proved}}$

In the figure given below , OL is perpendicular to AB and OM is perpendicular to BC such that OL = OM. Is $\Delta OLB \cong \Delta OMB$ ?

$Given,$

$OL = OM$

$\angle OLB$ =

$\angle OMB,$

$and$

$they$

$are$

$right$

$angles.$

$So,$

$OL^2 + BL^2 = OB^2$

$OM^2 + BM^2 = OB^2$

$OL^2 + BL^2$ =

$OM^2 + BM^2$

$BL^2$

$=$

$BM^2$

$\therefore$

$BL = BM$

$\angle OLB$ =

$\angle OMB$

$OL = OM$

$By$

$SAS$

$congruence,$

$\Delta OLB \cong \Delta OMB$

The lengths of the two sides of a triangle are $6$ cm and $10$ cm. Between what two wholes numbers should lie the measure of the third sides?

Two sides of triangle are

$6,10$

(Case 1) :-

let the third side be the smallest side

$x\leq 6$

$x+6>10$ {property of triangle}

$x>4$

$\therefore 4<x\leq 6$

$x\in (4,6]$

(Case 2) :-

Let the third side be the largest

$x\geqslant 10$

$10+6>x$ {property of triangle}

$x<16$

$x\in [10,16)$

$\therefore$

$x\in (4,16)$

$\therefore$

$x$ should lie in between

$4$ and

$16$ and not equal to

$4$ and

$16$ .

$ABCD$ is a trapzium in which $AB||CD$ and $AD=BC$ . Show that
$\angle A=\angle B$

1188077_1150650_ans_a841eccda4a4489fbdd9c10c945ea95c.JPG

$ABCD$ is a trapzium in which $AB||CD$ and $AD=BC$ . Show that $\triangle ABC \cong \triangle BAD$

Ref. image (I)

Given : ABCD is a trapezium where

$AB || CD$ and

$AD = BC$

To prove :

$\triangle ABC \cong \triangle BAD$

Construction : Extend

$AB$ and draw a line through

$C$ parallel to

$DA$ intersecting

$AB$ produced at

$E$

Proof :

$AD || CE$ (From construction) &

$AE || DC$ (As

$AB || DC, \ \& \ AB$ is extended)

$AECD$ , both pair of opposite sides are parallel,

$AECD$ is a parallelogram

$\therefore AD = CE$ (Opposite sides of parallelogram are equal) ......... Ref. image (III)

But

$AD = BC$ (Given)

$\Rightarrow BC = CE$

So,

$\angle CEB = \angle CBE$ (In

$\Delta BCE$ , Angles opposite to equal sides are equal) ...(1)

For

$AD || CE$ &

$AE$ is the transversal

$\angle A + \angle CEB = 180^{\circ}$ .........(Interior angle on same side of transversal is supplementary)

$\angle A = 180^{\circ} - \angle CEB$ ...(2)

Also

$AE$ is a line,

So,

$\angle B + \angle CBE = 180^{\circ}$ (Linear pair)

$\angle B + \angle CEB = 180^{\circ}$ (From (1))

$\angle B = 180^{\circ} - \angle CEB$ ...(3)

From (2) & (3)

$\angle A = \angle B$ ...............(4)

Now, Ref. image II

$\Delta ABC$ and

$\Delta BAD$

$AB = BA$ (common)

$\angle B = \angle A$ (Proved in part (i))

$BC = AD$ (Given)

$\therefore \Delta ABC \cong \Delta BAD$ (SAS congruence rule)

1430073_1150657_ans_98c0bf756c2f4588b6f04ab2b86d5ca6.png

$\angle C=\angle D.$

Let the line

$AB$ is extended up to point E such that

$AD||EC$

$AE||DC$

$AECD$ both pair of opposite sides parallel

$AECD$ is a parallelogram

$AD=CE$ (opposite side of parallelogram)

$AD=BC$

$\Rightarrow \ BC=CE$

$\angle CEB=\angle CBE$

$AD||CE$

$AE$ is the transversal

$\angle A+\angle CEB=180^o$

$\angle A=180^o -\angle CEB$

$(2)$ and

$(3)$

$\angle A=\angle B$

Similarly

$\angle C=\angle D$

Also

$AE$ is a line

$\angle B+\angle CBE=180^o$

$\angle B+\angle CEB=180^o$

$\angle B=180^o-\angle CEB$

1377250_1150653_ans_a71992a69e3748c189c89318c3e0a12d.png

PR is diagonal of a parallelogram PQRS. Show that $\Delta$ PQR $\cong$ $\Delta$ RSP.

Given that

$PQRS$ is a parallelogram.

Hence,

$PQ\parallel RS$ and

$PS\parallel QR$

We know that, alternate interior angles formed by a line intersecting two parallel lines, are equal.

$\therefore \ \angle SRP=\angle RPQ$ (Alternate interior angles)

$PR=PR$ (Common)

$\angle SPR=\angle PRQ$ (Alternate interior angles)

According to ASA congruency Criterion

$\triangle PQR\cong\triangle RSP\quad \quad [Hence\ proved]$

In the figure, ray $AZ$ bisects $\angle {DAB}$ as well as $\angle{DCB}$
(i) State the three pairs of equal parts in triangles $BAC$ and $DAC$ .
(ii) Is $\triangle {BAC}\cong \triangle {DAC}$ ? Give reasons
(iii) Is $AB=AD$ ? Justify your answer
(iv) Is $CD=CB$ ? Give reasons.

(i)

$\begin{array}{l} \angle DAC=\angle BAC \\ \angle DCA=\angle BCA \\ AC=AC \end{array}$

(ii)

By ASA criterion, triangles are congruent.

(iii)

Since triangles are congruent hence,

$AB=AD$ because both sides originate from equal angles in two triangles.

(iv)

Since triangles are congruent hence,

$CD=CB$ because both sides originate from equal angles in two triangles.

Check whether the following pairs of triangles are congruent. If they are congruent, state the congruence criterion.

$1).$ In

$\triangle ABC$ and

$\triangle PQR$

$AB=RQ=6$

$BC=QP=3$

$CA=PR=5$

$S.S.S$ criteria

$\triangle ABC\cong \triangle RQP$

$2)$ In

$\triangle ABD$ and

$\triangle BCD$

$AB=CD$

$BD=DB$

$DA=BC$

$S.S.S$ criteria

$\triangle ABD\cong \triangle CDB$

By applying $SAS$ congruence rule, you want to establish that $\Delta P Q R \cong \Delta F E D.$ It is given that $P Q = F E \text { and } R P = D F.$ What additional information is needed to establish the congruence ?

Since for

$\Delta PQR \cong \Delta FED$

Given

$PQ=FE,RP=DF$ then the additional required information will be the angle included

Additional information is required that

$\angle QPR$ must be equal to

$\angle EFD$

ABC is a triangle right angled at C. A line through the mid point M of hypotenuse AB and parallel to BC intersect AC at D .Show that :
$CM=MA=\dfrac{1}{2}AB$

$In\quad \Delta ABC$

$M$ is the midpoint of

$AB$

and

$\\ MD\parallel BC$

therefore

$D$ is the midpoint of

$AC$ [line drawn through midpoint of one side of a triangle parallel to another side, bisect the third side]

$\Rightarrow AD=DC\longrightarrow (1)$

also,

$\angle ADM=\angle CDM={ 90 }^{ o }\quad (corresponding\quad angle)\\ In\quad \Delta ADM\quad and\quad \Delta CDM\\ AD=CD\quad (from(1))\\ \angle ADM=\angle CDM={ 90 }^{ o }\\ DM=DM\quad (common)\\ \therefore \Delta ADM\cong \Delta CDM\quad (SAS)\\ \therefore AM=CM\quad (CPCT)\\ \because AM=\dfrac { 1 }{ 2 } AB\quad (M\quad is\quad the\quad midpoint\quad of\quad AB)\\ \therefore CM=MA=\dfrac { 1 }{ 2 } AB$

1170110_1184848_ans_09cccf06374c458aa3d310ccafcaf890.png

Find number of integral values of $\lambda$ if $\left( {\lambda ,\lambda + 1} \right)$ is an interior point of $\Delta ABC$ , where $A=(0,3), B=(-2,0)$ and $C=(6,1)$ .

Given,

$A=(0,3),B=(-2,0)$ and

$C=(6,1)$

Now, for the point to be an interior point, the basic conditions that can be applied are,

Condition

$1:$

$-2 < x < 6$

Condition

$2:$

$0 < y < 3$

For an interior point

$P(x,y)$

So, if

$\lambda=-1, (\lambda,\lambda+1)=(-1,0)$ , not an interior point.

$[Using\ condition\ 2]$

$\lambda=0, (\lambda,\lambda+1)=(0,1)$ , an interior point.

$\lambda=1, (\lambda,\lambda+1)=(1,2)$ , an interior point.

$\lambda=2, (\lambda,\lambda+1)=(2,3)$ , not an interior point.

$[Using\ condition\ 2]$

Ans so on . . . .

So, we have only

$2$ points

$(0,1)$ and

$(1,2)$ which will be inside the triangle

$ABC$

And thus

$\lambda=0,1.$

Hence

$\lambda$ can have

$2$ values.

1364944_1178328_ans_ebba8fd9ff94469a823dfdf8e5e6e89a.png

In the given figure, triangles $ABC$ and $DCB$ are right-angle at $A$ and $D$ respectively and $AC=DB$ . Prove that $\Delta ABC \cong \Delta DCB$

By RHS criteria both triangles are congruent

$\Delta ABC \cong \Delta DCB$

1222821_1235734_ans_4db3edf208884e1ca58d85764862b4ce.jpg

Express the congruence in the given triangles. Write in symbolic form.
In $\Delta XYZ,\ XY=4.2\ cm,\ XZ=6.5\ cm\ \angle Y=90^{o}$
In $\Delta DEF,\ FE=6.5\ cm,\ FD=4.2\ cm\ \angle D=90^{o}$

$\triangle XYZ$ and

$\triangle DEF$ ,

$XZ=FE=6.5 \ cm$

$XY=FD=4.2 \ cm$

$\angle Y=\angle D=90^{o}$

Hence,

$\triangle XYZ \cong \triangle DEF$ by

$RHS$ criterion.

In the figure $AB || DC$ and $AC$ and $PQ$ intersect each other at the point $O$ . Prove that $OA.CQ = OC.AP$ .

$\triangle APO$ and

$\triangle CQO$

$\angle AOP=\angle COQ$ (vertically opposite angle)

$\angle APO=\angle CQO$ (alternative angle)

$\angle PAO=\angle QCO$ (alternative angle)

Thus,

$\cfrac{OA}{AP}=\cfrac{OC}{CQ}$

So,

$OA.CQ=OC.AP$

Hence proved

By applying $SAS$ congruence rule, you want to establish that $\triangle PQR \equiv \triangle FED$ . It is given that $PQ=FE$ and $RP=DE$ . What additional information is needed to establish the congruence?

Given

$PQR \cong FED$

In SAS congruence rule two sides and one angle are equal to 2 sides & one included angle of another triangle.

Hence

$\angle P = \angle F$

And

$\angle Q \cong \angle E$ SAS means angle between

$2$ sides.

In parallelogram ABCD, two points P and Q are taken on diagonal BD such that $DP = BQ$ (refer to the given figure). Show that :
(i) $\triangle APD \cong \triangle CQB$
(ii) $AP=CQ$
(iii) $\triangle AQB \cong \triangle CPD$
(iv) $AQ=CP$
(v) APCQ is a parallelogram

In parallelogram

$ABCD$ , two points

$P$ and

$Q$ are taken on diagonal

$BD$ such that

$DP=BQ$

proof:(i)In

$\triangle APD$ and

$\triangle {CQB}$ ,

$\therefore AD\parallel BC$

opposite sides of parallelogram

$ABCD$ and a transversal

$BD$ intersect them

$\therefore \angle{ ABD}=\angle{ CBD}$ alternate interior angles

$\Rightarrow \angle ADP=\angle CBQ$ ....

$(1)$

$DP=BC$ ....

$(2)$

$AD=CB$ ......

$(3)$

opposite sides of parallelogram

$ABCD$

From

$\left(1\right) , \left(2\right)$ and

$\left(3\right)$

$\triangle APD\cong\triangle CQB$ by SAS congruence

(ii)

$\therefore \triangle APD =\triangle CQB$ proved in

$(1)$ above

$\therefore AP=CQ$

(iii)In

$\triangle AQB$ and

$\triangle CPD$

$\therefore AB||CD$ since opposite sides of parallelogram

$ABCD$ and a transversal

$BD$ intersects them

$\therefore \angle ABD=\angle CDB$ as alternate interior angles

$\Rightarrow \angle ABQ=\angle CDP$

$\Rightarrow QB=PD$ (given)

$\Rightarrow AB=CD$ opposite sides of parallelogram

$ABCD$

$\therefore \triangle {AQB}\cong\triangle{ CPD}$ by SAS congruence rule

$(iv)\therefore \triangle AQB=\triangle CPD$ proved in

$\left(iii\right)$ above

$\therefore AQ=CP$ by CPCT

$(v)\therefore$ the diagonals of a parallelogram bisects each other

$\therefore OB=OD$

$\therefore OB-BQ=OD-DP$

$\therefore BQ=DP$ (given)

$\therefore OQ=OP.....(1)$

Also,

$OA=OC.......(2)$

$\therefore$ diagonals of a parallelogram bisects each other

From

$(1)$ and

$(2), APCQ$ is a parallelogram

Diagonal AC of a parallelogram ABCD bisects $\angle A$ (see Fig. 8.19), show that
(i) it bisects $\angle C$ also (ii) ABCD is a rhombus

T.P (i)

$AC$ bisects

$C$

i.e.,

$\angle 3 = \angle 4$

(ii)

$ABCD$ is rhombus

Sol:

$\angle 1 = \angle 2$

$\angle 2 = \angle 3$ (alternate opposite)

$\angle 1 = \angle 4$

i.e.,

$\angle ACD = \angle CAD$

$AD = CD$ (opposite sides of equal angle in a triangle are equal)

but

$AB - CD$ and

$AD = BC$

$\Rightarrow AB = BC = CD = DA$

Thus

$ABCD$ is a rhombus

In $\Delta A B C$ $BE$ and $CF$ are the two equal altitudes. Prove $\Delta A B C$ is isoceles by R.H.S

$\begin{array}{l} Given \\ \angle AEB=\angle CEB={ 90^{ \circ } }\, \, \, \, \, \, \, ----\left( 1 \right) \\ Also,\, \, CF\, \, is\, \, a\, \, altitude, \\ So,\, \angle AFC=\angle BFC={ 90^{ \circ } }\, \, \, \, \, \, \, \, \, \, ---\left( 2 \right) \\ Also,\, BE=CF\, \, \, \, \, \, ---\left( 3 \right) \\ We\, have\, to\, prove\, that\, \Delta ABC\, is\, isoceles. \\ In\, \, \Delta BCF\, and\, \Delta CBE \\ \angle BFC=\angle CEB={ 90^{ \circ } }\, \, \\ FC=EB\, \, \, \left( { from3 } \right) \\ BC=CB\, \, \, \left( { common } \right) \\ \therefore \Delta BCF\, \cong \, \Delta CBE\, \, \, \, by\, RHS\, congruence\, rule \\ \therefore \angle FBC=\angle ECB \\ so,\, \angle ABC=\angle ACB \\ AB=AC \\ Hence,\, \Delta ABC\, is\, an\, isosceles\, triangle. \end{array}$
1234333_1217687_ans_c76660ca6d374a98a9baf09a5c51f1a9.PNG

1234333_1217687_ans_c76660ca6d374a98a9baf09a5c51f1a9.PNG

In the given figure, $AB=DC$ and $BD=CA$ . Prove that $\Delta ABC\cong DCB$ .

Given

$AB=CD$

$BD=CA$

$\triangle ABC$ &

$\triangle DCB$

$BC=BC$ (Common side)

$\therefore \$ By

$SSS$ congurency

$\triangle ABC\cong \triangle DCB$

1348653_1230287_ans_9f4adf52f51a45828c9fc428c66703b4.png

In figure, $AB=AC$ and $D$ is any point on $BC$ produced. Show that $AD>AB$ .

Given :

$AB=AC$

To prove :

$AD>AB$

Proof :

$\Delta ABC$ ,

$\angle ABC=\angle ACB=\theta$ (say) [as

$AB=AC\Rightarrow \Delta ABC$ is isoceles]

Now, we know that, in an isoceles triangle, the base angles are acute as only one obtuse angle can exist in a triangle

Now,

$\angle ACD={ 180 }^{ 0 }-\angle ACB$ (linear pair)

So, in

$\Delta ACD,\angle ACD$ is the largest angle

So, from triangle inequality, side opposite to the larger angle is greater than the side opposite to the smaller angle

So,

$AD>AC$

but,

$AC=AB$ (given)

$\Rightarrow AD>AB$

Hence proved.

1203540_1203503_ans_0697fd650b874ceaac8f60c3f3f6d526.jpg

State the correspondence between the sides and angels of the following congruent triangles.
a) $\Delta PQR \cong \Delta XYZ$
b) $\Delta ABC \cong \Delta TUV$

(a) Given that,

$\triangle PQR\text{ is congruent to }\triangle XYZ$

Therefore,

Congruent sides: $PQ=XY, \ QR=YZ \text{ and }PR=XZ$
Congruent angles: $\angle QPR=\angle YXZ, \ \angle PQR=\angle XYZ\text{ and } \angle PRQ=\angle XZY$

(b) Given that

$\triangle ABC \text{ is congruent to }\triangle TUV$

Therefore,

Congruent sides: $AB=TU, \ BC=UV\text{ and } AC=TV$
Congruent angles: $\angle BAC=\angle UTV,\ \angle ABC=\angle TUV\text{ and } \angle BCA=\angle UVT$

In a $\Delta A B C , A B = A C .$ If the bisectors of $\angle B$ and $\angle C$ meet $A C$ and $A B$ at points $D$ and $E$ respectively, show that :
(i) $\Delta \mathrm { DBC } \cong \Delta \mathrm { ECB }$
(ii) $\mathrm { BD } = \mathrm { CE }$

$AB=AC$

EC is bisector of

$\angle C$

BD is bisector of

$\angle B$

(i)

$AB=AC \rightarrow \angle B= \angle C$ (isosceler triangle)

$\Delta DBC$ and

$\Delta ECB$

$BC \rightarrow commonside$ ..(1)

$\angle DBC =1/2 \angle B = 1/2 \angle C$

$(\because \angle B=\angle C)$

$\angle ECB = 1/2 \angle B$

$\Rightarrow \angle DBC = \angle ECB$ ...(2)

$\angle EBC = \angle B = \angle C = \angle DBC$

$\Rightarrow \angle EBC = \angle DCB$ ...(3)

From (1), (2) & (3)

$\Delta DBC \cong \Delta ECB$

$\Rightarrow BD=CE$

$(\because \Delta DBC \cong ECB)$

1213143_1309896_ans_ad2b3bfbf67541a0bf2bef45798365a3.png

In figure, it is given that $AB=CD$ and $AD =BC$ . Prove that $\triangle ADC \cong \triangle CBA$ .

In $\Delta ADC$ and $\Delta CBA$

$AB=CD$ (Given)

$AD=BC$ (Given)

$CA=CA$ (common)

Hence $\Delta ADC \cong \Delta CBA$ by SSS congruence rule

1198310_1239568_ans_15f79aac39cc47fbabd5fc06659cce70.png

$\triangle EFG \cong \triangle LMN$
Write the corresponding vertices, angles and sides of the two triangles.

Given :

$\triangle EFG\cong \triangle LMN$

We are asked to find the corresponding vertices, angles sides of the two triangle.

We already know that the two triangles are congruent.

From the diagram, we can say that in

$\triangle EFG\;\&\; \triangle LMN$ :

$\Rightarrow FG=MN$

$EG=LN$

$EF=LM$

Hence, side corresponding to

$FG,\ EG\ and \ EF$ is

$MN,\ LN\ and\ LM$ respectively.

Now, again in

$\triangle EFG\;\&\; \triangle LMN$

$\angle FEG=\angle NLM$

$\angle EGF=\angle LNM$

$\angle EFG=\angle NML$

Hence, angle corresponding to

$\angle FEG,\ \angle EGF\ and \ \angle EFG$ is

$\angle NLM,\ \angle LNM\ and\ \angle NML$ respectively.

Hence, vertex corresponding to

$E,\ F\ and \ G$ is

$L,\ M\ and\ N$ respectively.

1211197_1243781_ans_8e2346f719734a27af33640e92ae3241.png

In the given figure, the triangles are congruent, Find the values of x and y.

We have,

According to given question,

$\Delta ABC\cong \Delta PQR$

Then,

$\angle ABC=\angle PQR$

$\Rightarrow x-{{10}^{o}}={{75}^{o}}$

$\Rightarrow x={{85}^{o}}$

And

$\angle ACB=\angle PRQ$

$\Rightarrow y-{{5}^{o}}={{25}^{o}}$

$\Rightarrow y={{30}^{o}}$

Hence, this is the answer.

In figure, two lines $AB$ and $CD$ intersect each other at the point $O$ such that $BC\parallel DA$ . Show that $O$ is the mid-point of both the line-segments $AB$ and $CD$ .

$\begin{array}{l} In\, \, \Delta COB\, \& \, \, \Delta AOD\, \left( { vertically\, \, opposite\, \, ange } \right) \\ \angle COB=\angle AOD\, \, \, \left( { alternate\, \, { { int } }erior\, \, angle } \right) \\ AB:BC\left( { given } \right) \\ \therefore \Delta COB\cong \Delta AOD\, \, \left( { ASA\, \, congurency } \right) \\ \therefore OB:OA\left( { CPCT } \right) \\ and\, \, OC=OD\left( { CPCT } \right) \\ and\, \, hence\, \, \, O\, \, is\; the\, \, mid\, \, po{ { int } }. \end{array}$

Name pair of congruent triangle in the figure.

Considering center point as

$O$

The pair of congruent trianlges are:

$\triangle AOD\cong \triangle BOC$

$\triangle AOB\cong \triangle COD$

$\triangle DAB\cong \triangle DCB$

$\triangle ADC\cong \triangle ABC$

$PSDA$ is a parallelogram. Points $Q$ and $R$ are takes on $PS$ such that $PQ=QR$ and $PA||QB||RC$ . Prove that area $(\Delta PQE)=(\Delta CFD)$

Consider

$\Delta PQE=\Delta DCF$

Since

$AD||PS$ and

$AP||BQ||CR||DS$

$\angle E=\angle F$

$EQ=FC\\PE=DF$

So By

$SAS$ criteria

$\Delta PQE$ is congruent to

$\Delta CFD$

$ar(\Delta PQE)=ar(\Delta CFD)$

in the adjacent figure , parallelogram $ABCD$ , Show that $ABC \cong CDA$

$\begin{array}{l} In\, \, \Delta ABC\, \, \& \, \, \Delta CDA \\ AB=CD\, \, \\ BC=DA \end{array}$

(opposite side of parallelogram are equal)

$\angle ABC=\angle CDA$ Oppsite angle in parallelogram are equal

$\therefore \Delta ABC\cong \Delta CDA\, \, \left( { SAS\, \, congurence\, \, rule } \right)$

Prove that in an isosceles triangle, the angles opposite the equal sides are equal.

R.E.F. Image.

Given ABC is an isosceles triangle.

proof :- Given :- AB = AC.

To prove :-

$\angle B=\angle C.$

Let AD is angle bisector of angle A.

and D be point of interactions of this bisector

$\angle A$ & BC

$\triangle BAD$ &

$\triangle CAD$

$AB = AC$ (Given)

$\angle BAD = \angle CAD$ (By constriction)

$AD = AD$ (common)

$\triangle BAD \cong \triangle CAD$ [By SAS rule]

$\angle ABD = \angle ACD$ [CPCT]

$\therefore [\angle B = \angle C]$

1206716_1277283_ans_1c6081c2f1c941ea87086b2d8caca7fe.JPG

$\triangle{ABC}$ and $\triangle{DBC}$ are two isosceles triangles on the same base $BC$ and vertices $A$ and $D$ are on the same side of $BC$ as in figure.If $AD$ is extended to intersect $BC$ at $P,$ Show that $\triangle{ABP}\cong\triangle{ACP}$

Given:

$\triangle{ABC}$ is isosceles,

$AB=AC$ .....

$(1)$

Also,

$\triangle{DBC}$ is isosceles,

$DB=DC$ .....

$(2)$

Proof:In

$\triangle{ABD}$ and

$\triangle{DBC}$ , we have

$AB=AC$ from

$(1)$

$BD=DC$ from

$(2)$

$AD=AD$ (common)

$\triangle{ABD}\cong\triangle{ACD}$ by SSS congruence rule.

So,

$\angle{BAP}=\angle{PAC}$ by CPCT .........

$(3)$

$\triangle{ABP}$ and

$\triangle{ACP},$

$AB=AC$ (given)

$\angle{BAP}=\angle{PAC}$ by CPCT (from

$(3)$ )

$AP=AP$ (common)

$\triangle{ABP}\cong\triangle{ACP}$ by SAS congruence rule.

If in two right triangles the hypotenuse and one side of one triangle are equal to the hypotenuse and one side of the other triangle, then prove that two triangles are congruent.

Given:Two right triangles

$\triangle{ABC}$ and

$\triangle{DEF}$ where

$\angle{B}={90}^{\circ}$ and

$\angle{E}={90}^{\circ}$ , hypotenuse is equal.

i.e.,

$AC=DF$ and one side is equal i.e.,

$BC=EF$

To prove:

$\triangle{ABC}\cong\triangle{DEF}$

Proof:In right

$\triangle{ABC}$

By pythagoras theorem,

${AC}^{2}={AB}^{2}+{BC}^{2}$

${AB}^{2}={AC}^{2}-{BC}^{2}$ ........

$(1)$

In right

$\triangle{DEF}$

By pythagoras theorem,

${DF}^{2}={DE}^{2}+{EF}^{2}$

${DE}^{2}={DF}^{2}-{EF}^{2}$ ........

$(2)$

From

$(1)$

${AB}^{2}={AC}^{2}-{BC}^{2}$

$\Rightarrow {AB}^{2}={DF}^{2}-{EF}^{2}$ since

$AC=DF$ and

$BC=EF$

$\Rightarrow {AB}^{2}={DE}^{2}$ from

$(2)$

$\Rightarrow AB=DE$ ......

$(3)$

Given:

$BC=EF$ ,

$AC=DF$

$\Rightarrow \triangle{ABC}\cong\triangle{DEF}$ by SSS congruence.

Hence proved.

1264284_1377826_ans_5e41dfa19d7144f386aa72e4db5262cf.PNG

In $\triangle{PQR},N$ is a point on $PR$ such that $QN\bot PR$ . If $PN\times NR={QN}^{2}$ , prove that $\angle{PQR}={90}^{\circ}$

$PN\times NR=QN^2$

Given:

$PN=QN, QN=NR$

$\Delta PNQ$ and

$\Delta QNR$

$\angle PNQ=\angle QNR=90^o$

$PN=QN, QN=NR$

$\Delta PNQ\sim \Delta QNR$ by S.A.S

$\therefore \angle PQN=\angle QRN$

$\Delta QNR$ ,

$\angle QNR+\angle QRN+\angle NQR=180^o$ by angle sum property

$\Rightarrow 90^o+\angle PQN+\angle NQR=180^o$

$\Rightarrow \angle PQN+\angle NQR=180^o-90^o=90^o$

$\Rightarrow\angle PQR=90^o$

Hence proved.

1222844_1319840_ans_4594e96327d34d18a19883efd02c4723.PNG

$\triangle{ABC}$ and $\triangle{DBC}$ are two isosceles triangles on the same base $BC$ and vertices $A$ and $D$ are on the same side of $BC$ as in figure.If $AD$ is extended to intersect $BC$ at $P,$ Show that $\triangle{ABD}\cong\triangle{ACD}$

Given:

$\triangle{ABC}$ is isosceles,

$AB=AC$ .....

$(1)$

Also,

$\triangle{DBC}$ is isosceles,

$DB=DC$ .....

$(2)$

Proof:In

$\triangle{ABD}$ and

$\triangle{ACD}$ , we have

$AB=AC$ from

$(1)$

$BD=DC$ from

$(2)$

$AD=AD$ (common)

$\triangle{ABD}\cong\triangle{ACD}$ by SSS congruence rule.

$ABCD$ is a square and $BD$ is the diagonal. Find the measure of the area of $\triangle ABD$ .

$\triangle{ABD}$ and

$\triangle{BCD}$ ,

$AB = CD \quad \left( \text{Each sides of a square are equal} \right)$

$\angle{BAD} = \angle{BCD} \quad \left( \text{Each } 90° \right)$

$AD = BC \quad \left( \text{Each side of a square are equal} \right)$

$\therefore \triangle{ABD} \cong \triangle{BCD} \quad \left( \text{By SAS} \right)$

Therefore,

$ar{\left( \triangle{ABD} \right)} = ar{\left( \triangle{BCD} \right)} \quad \left( \text{Congruent triangles} \right)$

Therefore,

Area of square

$= ar{\left( \triangle{ABD} \right)} + ar{\left( \triangle{BCD} \right)}$

$\Rightarrow 2 ar{\left( \triangle{ABD} \right)} =$ Area of square

$\Rightarrow ar{\left( \triangle{ABD} \right)} = \cfrac{1}{2}$ Area of square

If area of $\triangle{ABC}=$ area of $\triangle{DEF}$ then prove that $\triangle{ABC}\cong\triangle{DEF}$

Given

$\triangle{ABC}\sim\triangle{DEF}$

We know that if two triangles are similar, then the ratio of there are equal to the square of the ratio of its corresponding sides.

$\dfrac{area{\triangle{ABC}}}{area{\triangle{DEF}}}={\left(\dfrac{BC}{EF}\right)}^{2}={\left(\dfrac{AB}{DE}\right)}^{2}={\left(\dfrac{AC}{DF}\right)}^{2}$

$\dfrac{area{\triangle{DEF}}}{area{\triangle{DEF}}}={\left(\dfrac{BC}{EF}\right)}^{2}={\left(\dfrac{AB}{DE}\right)}^{2}={\left(\dfrac{AC}{DF}\right)}^{2}$

since

$area{\triangle{ABC}}=area{\triangle{DEF}}$

$\Rightarrow 1={\left(\dfrac{BC}{EF}\right)}^{2}={\left(\dfrac{AB}{DE}\right)}^{2}={\left(\dfrac{AC}{DF}\right)}^{2}$

$\Rightarrow 1={\left(\dfrac{BC}{EF}\right)}^{2},\,\,\,\,1={\left(\dfrac{AB}{DE}\right)}^{2},\,\,\,\,1{\left(\dfrac{AC}{DF}\right)}^{2}$

$\Rightarrow \dfrac{BC}{EF}=1,\,\,\,\,\dfrac{AB}{DE}=1,\,\,\,\,\dfrac{AC}{DF}=1$

$\Rightarrow BC=EF,\,\,\,\,AB=DE,\,\,\,\,AC=DF$

Hence by SSS congruency,

$\triangle{ABC}\cong\triangle{DEF}$

Hence proved.

In the given figure $AB=CD$ and $AD=BC$ . Prove that $\angle BAC= \angle ACD$ .

$\Delta ABC$ and

$\Delta CDA$ :

$AC = AC$ (common)

$AB = CD$ (Given)

$AD = BC$ (Given)

$SSS$ property:

$\Delta$ ABC

$\cong \Delta$ CDA

Thus,

$\angle$ BAC =

$\angle$ ACD

Hence, proved.

$\triangle ABC$ is a right angled in which $LC=90^{o}$ and $CD\bot AB$ . If $BC=a, CA=b, AB=c$ and $CD=p$ then prove that
$cp=ab$

Area of triangle is

$base \times height$

When

$c$ is taken as base Then,

$p$ becomes height so the area is

$\dfrac 12cp$

When

$b$ is taken as base Then,

$a$ becomes height so the area is

$\dfrac 12ab$

Since area is same

$\implies ab=cp$

Is it possible to have a triangle with the following sides:
$3\ cm,\ 4\ cm$ and $5\ cm$

Sum of any two sides of triangle should be greater than the third side.

Here,

$3+4 > 5$

$4+5 >3$

$3+5 >4$

Hence, it is a possible triangle.

$\triangle ABC$ is a right angled in which $LC=90^{o}$ and $CD\bot AB$ . If $BC=a, CA=b, AB=c$ and $CD=p$ then prove that
$\dfrac{1}{p^{2}}=\dfrac{1}{a^{2}}+\dfrac{1}{b^{2}}$

Area of triangle is

$base \times height$

When

$c$ is taken as base Then,

$p$ becomes height so the area is

$\dfrac 12cp$

When

$b$ is taken as base Then,

$a$ becomes height so the area is

$\dfrac 12ab$

Since area is same

$\implies ab=cp$

$\dfrac 1{a^2}+\dfrac 1{b^2}\\\dfrac{a^2+b^2}{(ab)^2}\\\dfrac{c^2}{c^2p^2}=\dfrac 1{p^2}$

By applying ASA congruence rule, it is to be established that $\Delta ABC \cong \Delta QRP$ and it is given that BC $=$ RP. What additional information is needed to establish the congruence?

Given

$BC= RP$

And

$\Delta ABC \cong \Delta QPR$ (By ASA congruence rule)

Since it is ASA congruence rule,

we need

Angle

Side

Angle

Here, side is

$BC=RP$ .

So, Angle on other side of it in both of those triangles should be equal

So,

$\angle A=\angle Q$

Therefore, for another angle we need additional information on the following conditions:

$\angle B=\angle R$

$\angle C=\angle P$

1201533_1416413_ans_20872e2a18e04db6a7186a413f8a84af.png

AB and CD intersect each other at O and O is the midpoint of both AB and CD. Prove that AD=BC.

Given

$O$ is the mid point of

$AB$ and

$CD$

$\implies OA=OB,OC=OD$

$\triangle AOD,\triangle BOC$

$AO=BO,CO=DO,\angle AOD=\angle BOC (\mathrm{vertically\ opposite\ angles})$

$S.A.S$ congruency

$\triangle AOD\cong\triangle BOC$

$\implies AD=BC$

In the following figure, state the conditions you would use to show that $\Delta ABC$ and $\Delta EDC$ are congruent.

We have to find

$\Delta$ ABC and

$\Delta$ EDC are congruent

$\Delta$ ACB and

$\Delta$ EDC

$\angle ACB=\angle ECD$ (vertically opposite angles)

$CE=AC$ (given)

$CB=CD$ (given)

By SAS property.

$\Delta ACB\cong \Delta EDC$ .

1221395_1450245_ans_7789a87dfcee4e7ba8fae245bd4570a2.png

In the given figure, it is given that $AB=CD$ and $AD=BC$ . Prove that $\Delta ADC \cong \Delta CBA$

$\Delta$ ACD &

$\Delta$ CBA

$AB=CD$ (Given)

$AD=BC$ (Given)

$AC=AC$ (common)

By SSS

$\Delta ADC\cong \Delta CBA$ .

In given figure explain, why $\triangle ABC\cong \triangle FED$

Given:-

$\angle{BAC} = \angle{FED}$ and

$BC = DE$

To prove:-

$\triangle{ABC} \cong \triangle{FED}$

Proof:-

$\triangle{ABC}$ and

$\triangle{FED}$

$\angle{BAC} = \angle{EFD} \quad \left( \text{Given} \right)$

$\angle{ABC} = \angle{DEF} \quad \left( \text{Each } 90 \right)$

$BC = DE \quad \left( \text{Given} \right)$

$\therefore \triangle{ABC} \cong \triangle{FED} \quad \left( \text{By AAS congruence rule} \right)$

Hence proved.

In the adjoining figure, sides $AB$ and $AC$ of $\triangle{ABC}$ are extended to points $P$ and $Q$ respectively.Also, $\angle{PBC} < \angle{QCB}$ . Show that $AC>AB.$

REF.Image

Given

$\angle PBC < \angle QCB$

$\angle PBC$ is the external angle of

$\Delta ABC$

so,

$\angle PBC=\angle A+\angle ACB$

Now,

$\angle PBC < \angle QCB$

$\Rightarrow \angle A+\angle ACB < \angle A+\angle ABC$

$\Rightarrow \angle ACB < \angle ABC$

$\Rightarrow AB < AC (\because$ Side opposite to the greater angle is longer)

Thus

AC > AB

Hence Proved

1219011_1505887_ans_2adc054202f8462d925803a63b232899.png

Take any point $O$ in the interior of a triangle $PQR$ . Is $OQ+OR>QR?$

Join

$O$ with

$P, Q, R$

Now in a

$\Delta le$

Sum of 2 side is greater than 3rd side

$\therefore OQ + OR > QR$

As,

$OQR$ is a triangle.

1339399_1645773_ans_07e348ded89f457cb8c7c4295b8ff7a1.png

Is it possible to have a triangle with the following sides?
$6$ cm, $3$ cm, $2$ cm.

No. Because the triangle has the property that sum of any two sides is always greater than the third side but in this case it is not possible because

$3+2=5cm$ and it is not greater than the third side that is

$6cm$ .

Take any point O in the interior of a $\triangle PQR$ . Is $OP$ $+$ $OQ$ $>$ $PQ$ $?$

Join

$O$ with

$P, Q, R$

As,

$OPQ$ is a triangle.

Sum of two sides is always greater than the third side.

$\therefore OP + OQ > PQ$

1339398_1645772_ans_cc4e4119e143478a8ab4ef1df44bff1a.png

Take any point $O$ in the interior of a triangle $PQR$ . Is $OR+OP>RP?$

Join

$O$ with

$P, Q, R$

As,

$OPR$ is a triangle.

Sum of 2 side is greater than 3rd side

$\therefore OP + OR > PR$

1339401_1645774_ans_cd09308b564d4ff39e4eaf5e1025a849.png

Is it possible to have a triangle with the following sides?
$3$ cm, $6$ cm, $7$ cm.

The length of a side of a triangle is less than the sum of the length of other two sides and greater than the difference of length of 2 side

Given,

$3cm, 6cm, 7cm$

$3 + 6 > 7$

$7 - 6 < 3$

$3 + 7 > 6$

$7 - 3 > 6$

$6 + 7 > 3$

$6 - 3 < 7$

$\therefore$ Possible

In the given figure, $AB=AD,\,\angle 1=\angle 2=\angle 3=\angle 4$ . Prove that $AP=AQ$ .

Given is a figure where

$AB=AD,\ \angle 1=\angle 2=\angle 3=\angle 4$

To prove:

$AP=PQ$

Proof:-

$AD=AB$ (Given)

$\angle 3=\angle 4$ (Given)

$\angle 1=\angle 2$ (Given)

$\therefore \ \angle 3+\angle 1=\angle 4+\angle 2$

Hence,

$\angle BAC=\angle DAC$

$AC=AC$ [common]

$\therefore \ \Delta ABC\cong \Delta ADC$ (SAS rule)

Hence,

$\angle BCA=\angle DCA\quad (CPCT)$

Now, in

$\Delta APC \ and \ \Delta AQC$ ,

$\angle 3=\angle 4$ (Given)

$AC=AC$ [common]

and

$\angle PCA=\angle QCA\quad \quad [\because \ \angle BCA=\angle DCA]$

Hence,

$\Delta APC\cong \Delta AQC$ [ASA rule]

$\therefore AP=AQ\quad [CPCT].\quad \quad [Hence\ proved]$

Is it possible to have a triangle with the following sides?
$1$ cm, $3$ cm, $5$ cm.

No. Because the triangle has the property that sum of any two sides is always greater than the third side but in this case it is not possible

$3+1=4cm$ and it is not than the third side that is

$6cm$

In the given figure, $AB\parallel CD$ and $O$ is the midpoint of $AD$ .
Show that (i) $\triangle AOB\cong \triangle DOC$
(ii) $O$ is the midpoint of $BC$ .

(i) From the figure,

$\triangle AOB$ and

$\triangle DOC$

We know that

$AB \parallel CD$ and

$\angle BAO$ and

$\angle CDO$ are alternate angles

So we get

$\angle BAO = \angle CDO$

From the figure, we also know that

$O$ is the midpoint of the line

$AD$

We can write it as

$AO = DO$

According to the figure we know that

$\angle AOB$ and

$\angle DOC$ are vertically opposite angles.

So we get

$\angle AOB = \angle DOC$

Therefore, by

$ASA$ congruence criterion we get

$\triangle AOB \cong \triangle DOC$ .

(ii) We know that

$\triangle AOB\cong \triangle DOC$

So we can write it as

$BO = CO (c. p. c. t)$

Therefore, it is proved that

$O$ is the midpoint of

$BC$ .

The lengths of the two sides of a triangle are $12$ cm and $15$ cm. Between what two measures should the length of the third side fall?

Since, the sum of lengths of any two sides in a triangle should be greater than the length of the third side.
It is given that two sides of triangle are

$12$ cm and

$15$ cm.
Therefore, the third side should be less than

$12+15=27$ cm.
And also the third side cannot be less than the difference of the two sides.
Therefore, the third side has to be more than

$15-12=3$ cm
Therefore, the third side should be the length more than

$3$ cm and less than

$27$ cm.

therefore,

$3cm<$ 3rd side

$<27cm$

In Fig. $AD \perp CD$ and $CB \perp CD.$ If $AQ = BP$ and $DP = CQ$ , prove that $\angle DAQ = \angle CBP.$

$\Delta 's ADQ$ and

$BCP,$ we have

$DQ = PC$ [

$\because DP = QC (Given)\therefore DP+PQ = PQ+QC\Rightarrow DQ = PC]$

$\angle ADQ=\angle BCP=90^{\circ}$ (given)

and,

$AQ = BP$ [Given]

So, by

$RHS$ congruence criterion, we obtain

$\Delta ADQ \cong \Delta BCP$

$\Rightarrow \angle A = \angle B$ i.e.,

$\angle DAQ = \angle CBP$ . (By C.P.C.T)

In a rectangle $ABCD$ , prove that $\triangle ACB \cong \triangle CAD$ .

$\triangle\text{ACB}$ and

$\triangle\text{CAD}$

$\text{AB}=\text{CD}$ [Opposite sides of a rectangle are equal]

$\text{AD}=\text{BC}$ [Opposite sides of a rectangle are equal]

$\text{AC}=\text{AC}$ [Common side]

Hence, by

$\text{SSS}$ rule of congruence,

$\triangle\text{ACB}\cong\triangle\text{CAD}$

1884873_1672986_ans_8b1dc15c392a46f092bd52ea03d587b1.png

Explain the following term.
Interior of a triangle.

Interior of a triangle is a set of all points lying inside of a triangle.

In above figure. shaded region inside

$\Delta ABC$ is an interior of

$\Delta ABC$ .

1616674_1677747_ans_77e366e238864fa6a2dbaa15e1d9e1c6.png

In Fig., there is a triangle. The measures of some angles have been indicated. State whether triangle is acute, right or obtuse.

An obtuse triangle is a triangle in which one angle is obtuse (more than 90°).

one of the angle(120 degrees) is greater than 90 degrees. so triangle is obtuse.

In Fig., there is a triangle. The measures of some angles have been indicated. State whether triangle is acute, right or obtuse.

An acute triangle is a triangle in which all the angles are acute (less than 90°).

All angles of triangle are less than 90 degrees. so triangle formed is acute.

In $\Delta ABC,$ side AB is produced to D so that BD = BC. If $\angle B = 60^{\circ}$ and $\angle A = 70^{\circ},$ prove that: $AD > AC$

We have,

$\angle A = 70^{\circ}$ and

$\angle B = 60^{\circ}.$

So,

$\angle C = 50^{\circ}; \angle CBD = 120^{\circ}$ and

$\angle BDC = \angle DCB = 30^{\circ}$

Now,

$\angle ACD = 50^{\circ}+30^{\circ} = 80^{\circ}, \angle CAD = 70^{\circ}$ and

$\angle ADC = 30^{\circ}$

$\therefore \angle ACD > \angle CAD$ and

$\angle ACD > CDA$

$\Rightarrow AD > CD$ and

$AD > AC$

The angles of a triangle are arranged in ascending order of magnitude. If the difference between two consecutive angles is $10^{\circ},$ find the three angles.

Let the angles be

$x^{\circ}, \left ( x+10 \right )^{\circ}$ and

$\left ( x+20 \right )^{\circ}.$ Then,

$x+\left ( x+10 \right )+\left ( x+20 \right ) = 180 \Rightarrow 3x+30 =180\Rightarrow 3x = 150\Rightarrow x = 50$
So, the angles are

$50^{\circ}, 60^{\circ},$ and

$70^{\circ}.$

In the given figure, it is given that $RT = TS, \angle 1 = 2\angle 2$ and $\angle 4 = 2\angle 3.$ Prove that $\Delta RBT \cong \Delta SAT.$

$\Delta RTS,$ we have

$RT = ST$

$\Rightarrow \angle TSR = \angle TRS$ (i)

We have,

$\angle 1= \angle 4$ [Vertically opposite angles]

$\Rightarrow 2\angle 2 = 2\angle 3$ [

$\because \angle 1 = 2\angle 2$ and

$\angle 4 = 2\angle 3$ (given)]

$\Rightarrow \angle 2 = \angle 3$ (ii)

Subtracting (ii) from (i), we get

$\angle TRS - \angle 2 = \angle TSR - \angle 3$

$\Rightarrow \angle TRB = \angle TSA$ (iii)

Thus, in triangles,

$RBT$ and

$SAT,$ we have

$\angle RTB = \angle STA$ [Common]

$RT = ST$ [Given]

and,

$\angle TRB = \angle TSA$ [From (iii)]

So, by ASA congruence criterion, we obtain

$\Delta RBT \cong \Delta SAT$

Explain the following term.
Obtuse triangle.

Obtuse triangle is a triangle in which one of the three angles is greater than

$90^0$ .

In above figure

$\Delta ABC$ is an obtuse angle because

$\angle B>90^0$ .

1616661_1677746_ans_ccfc64398f824799ac913a9deec62f4a.png

Fill in the blank:
Two squares are congruent if ________.

We know that, two geometrical figures are congruent if they have same shape and size.

Hence, two squares are congruent if they have the same side length.

Fill in the blank.
Two line segments are congruent if ________.

They are of equal lengths.

A line segment is defined by the distance between two points and two line segments will be congruent if both are of same lengths.

In the following triangle, the lengths of the sides are indicated along sides. By applying SSS condition, determine which are congruent. State the result in symbolic form.

1615310_1678054_ans_e4fa799508d64c1093364316eb554cb0.jpg

Explain the concept of congruence of figures with the help of certain examples.

1615135_1678018_ans_c9b085220cf5449b973dfef882fc2859.jpg

In the given figure, AB $=$ DC and BC $=$ AD. Is $\Delta$ ABC $\cong\Delta$ CDA?

$\Delta ABC$ and

$\Delta CDA$ ,

$AB = DC$ (given)

$BC = AD$ (given)

$AC = AC$ (common)

$\therefore \Delta ABC \cong \Delta CDA$ {By SSS cong. rule}

Answer is Yes.

2064094_1678059_ans_ff0b0fe23c72455498180c3dfa4f7118.png

Identify the pairs of triangles are congruent by ASA condition?

1616421_1678101_ans_daaaa96b3e004635a232d37e1c7addbb.jpg

Triangles ABC and PQR are both isosceles triangles with AB $=$ AC and PQ $=$ PR respectively. If AB $=$ PQ and BC $=$ QR, then are the two triangles congruent? Which condition do you use? If $\angle B=50^o$ , what is the measure of $\angle R$ ?

1615723_1678064_ans_e954ec1eb1264a428487e66d8e1dcf1a.jpg

In the following pairs of triangles, the lengths of the sides are indicated along sides. By applying SSS condition, determine which are congruent. State the result in symbolic form.

1615280_1678032_ans_0326ff3828554e07994a3d4e6bd3cf3f.jpg

In the given figure, line segments $AB$ and $CD$ bisect each other at $O$ . Prove that, $\Delta AOC\cong \Delta BOD$ .

Given that,

line segments

$AB$ and

$CD$ bisect each other at

$O$ .

So,

$O$ is the midpoint of

$AB$ and

$CD$

$\therefore OA=OB........(1)$

and,

$OC=OD........(2)$

Now, in

$\triangle AOC$ and

$\triangle BOD$

$OA=OB$ [from

$(1)$ ]

$OC=OD$ [from

$(2)$ ]

and,

$\angle AOC=\angle BOD$ [Vertically opposite angles]

Hence, by

$SAS$ congruency criterion we can say that,

$\Delta AOC\cong \Delta BOD$

Hence,

$proved$ .

In the given figure, $AB=AD$ and $\angle BAC=\angle DAC$ . Complete the following, so as to make it true:
Line segment $AC$ bisects _ and ___.

1616289_1678092_ans_9869e51ff4bd456ab2a0db2dbb7ede98.jpg

In figure, AD bisects $\angle$ A and AD $\perp$ BC. Is it true to say that BD $=$ DC?

As,

$AD$ is bisector of

$\angle A$ ,

$\Rightarrow \angle BAD\cong\angle CAD$

$\angle ADB\cong\angle ADC=90^o\cdots\cdots[\because\text{it is given that } AD\perp BC]$

And

$AD=AD\cdots\cdots[\text{common side}]$

So, by

$ASA$ test

$\Delta ABD\cong \Delta ACD$ .

Corresponding sides of congruent triangles are equal.

So,

$BD=CD$

Draw any triangle ABC. Use ASA condition to construct another triangle congruent to it.

Here, In

$ABC$ ,

$\angle B=60^o$ and

$\angle C=55^o$

$\Delta DCB$ is costructed on base

$BC$ such that

$\angle DCB=\angle ABC=60^0$

$BC=CB\cdots\cdots[\text{common side}]$

And

$\angle DBC=\angle ACB=55^o$

So, by

$ASA$ condition

$\Delta ABC\cong\Delta DCB$

1682674_1678108_ans_6f88c93b050f43239a13db0a67e47b1b.png

Identify the pairs of triangles are congruent by ASA condition?

In given figure ,

$\angle ADB=\angle ADC=90^o$ ,

$\angle DAC=50^o$ and

$\angle ABD=40^o$

$\Delta ABD$

$\angle ABD+\angle BDA+\angle DAB=180^o$

$\Rightarrow 40^o+90^0+\angle DAB$ =180^0$$

$\Rightarrow \angle DAB=50^o$

So, in

$\Delta ABD$ and

$\Delta ACD$

$\angle DAB\cong\angle DAC=50^o$

$AD=AD\cdots\cdots[\text{common side}]$

$\angle ADB\cong\angle ADC=90^o$

Hence, by

$ASA$ condition

$\Delta ADB\cong\Delta ADC$

In figure, AO $=$ OB and $\angle$ A $=\angle$ B. Is $\Delta$ AOC $\cong\Delta$ BOD?

It is given that

$AO$ =

$OB$ and

$\angle$

$A$

$=\angle$

$B$

Also,

$\angle AOC=\angle BOD\cdots[\text{vertically opposite angles}]$

So, by

$ASA$ test,

$\Delta$

$AOC$

$\cong\Delta$

$BOD$

State the three pairs of matching parts you have used to answer.(Is $\Delta$ ABC $\cong\Delta$ ACB)

$\Delta$ ABC, it is known that

$\angle B=\angle C$ ,

$\Delta$ ABC is isosceles

So,

$AB$ =

$AC$ ........

$(1)$ (Because opposite sides of equal angles are equal.)

Now, we have another copy of

$\Delta ABC$ which is .

$\Delta ACB$

$\Delta ABC\ \&\ \Delta ACB$

$(a)\ AB=AC\cdots\cdots[\text{from (1)}]$

$(b)\ BC=CB\cdots\cdots[\text{equal side}]$

$(c)\ AC=AB\cdots\cdots[\text{from (1)}]$

So, by

$SSS$ condition of congruence

$\Delta ABC\cong \Delta ACB$ is true.

Above written three pairs of equal sides are three pairs of of matching parts we have used to reach the answer.

In figure, AD bisects $\angle$ A and AD $\perp$ BC. State the three pairs of matching parts you have used after proving the triangles congruent.

$\Delta ABD$ and

$\Delta ACD$

$\angle BAD = \angle CAD$ (

$\therefore AD$ bisect

$\angle A$ )

$AD = AD$ (common)

$\angle ADB = \angle ADC (AD \bot BC)$

$\therefore \triangle ABD \cong \triangle ACD$ (ASA)

Thus the matching pairs,

$BD = DC$ (c.p.c.t)

$AB = AC$ (c.p.c.t)

$\angle ABC = \angle ACD$ (c.p.c.t)

2057430_1678106_ans_deb7a448eb0742d29175abc20fa9d529.png

Can the pairs of triangles be congruent by ASA condition?

In given triangles

$BC=QR=5\ cm$

But, their angles are not equal

So,

$\Delta ABC$ and

$\Delta PQR$ cannot be congruent by any condition

Is $\Delta$ ABC $\cong\Delta$ ACB?

$\Delta$ ABC, it is known that

$\angle B=\angle C$ , so

$\Delta$ ABC is isosceles

So,

$AB$ =

$AC$ ...................

$(1)$

Because side opposite of equal angles.

Now, we have another copy of

$\Delta$ ABC i.e.

$\Delta ACB$

So, in

$\Delta$ ABC

$\ and\ \Delta$ ACB

$(a)\ AB=AC\cdots\cdots[\text{from (1)}]$

$(b)\ BC=CB\cdots\cdots[\text{equal side}]$

$(c)\ AC=AB\cdots\cdots[\text{from (1)}]$

So, by

$SSS$ condition of congruence

$\Delta$ ABC

$\cong\Delta$ ACB is true.

Identify the pairs of triangles are congruent by ASA condition?

1616491_1678103_ans_1bf71d8c6cf84a20be7505a3b0bb88e7.jpg

In figure, AX bisects $\angle$ BAC as well as $\angle$ BDC. State the three facts needed to ensure that $\Delta$ ABD $\cong\Delta$ ACD.

AX bisects

$\angle$ BAC as well as

$\angle$ BDC.

So,

$\angle CAD=\angle BAD$

also

$\angle CDA=\angle BDA$

and

$AD=AD\cdots[\text{common side}]$

So, by

$ASA$ test,

$\Delta$ ABD

$\cong\Delta$ ACD.

Above written

$3$ conditions ensure that

$\Delta$ ABD

$\cong\Delta$ ACD.

In figure, BD and CE are altitudes of $\Delta$ ABC and BD $=$ CE. Is $\Delta$ BCD $\cong\Delta$ CBE?

By RHS congruence condition if hypotenuse and one side of one right triangle are equal to hypotenuse and one side of other right triangle then both right triangles are congruent.

It is given that

$BD=CE$ and

$BC$ is the hypotenuse of small triangles

$CEB$ and

$BDC$

So, In right triangles

$BCD$ and

$CBE$

$BC=CB=\text{common side}\cdots\cdots[\text{hypotenuse}]$

And

$BD=CE=\text{given}\cdots\cdots[\text{sides of right triangles}]$

So, by

$RHS$ condition

$\triangle BCD\cong\triangle CBE$

1683619_1678129_ans_eab759e5684747fe861b4e03c8eb3ca1.png

In the following pairs of right triangles, the measures of some parts are indicated along side. State by the application of RHS congruence condition which are congruent. State the result in symbolic form.

By RHS congruence condition if hypotenuse and one side of one right triangle are equal to hypotenuse and one side of other right triangle then both right triangles are congruent.

In right triangles

$ABO$ and

$DCO$

$AO=DO=4\ cm\cdots\cdots[\text{hypotenuse}]$

And

$BO=CO=3\ cm\cdots\cdots[\text{sides of right triangles}]$

So, by

$RHS$ condition

$\triangle ABO\cong\triangle DCO$

In figure, $AO=OB$ and $\angle A=\angle B$ . State the matching pair you have used for $\Delta$ $AOC$ $\cong\Delta$ $BOD$ , which is not given in the question.

It is given that

$AO=OB$ and

$\angle$

$A$

$=\angle$

$B$

Also,

$\angle AOC=\angle BOD\cdots[\text{vertically opposite angles}]$

So, by

$A.S.A.$ test,

$\Delta$

$AOC$

$\cong\Delta$

$BOD$

So, matching pair used which is not given in question is

$\angle AOC=\angle BOD$ which are opposite angles

In figure, AO $=$ OB and $\angle A=\angle B$ . Is it true to say that $\angle ACO=\angle BDO$ ?

It is given that

$AO$ =

$OB$ and

$\angle$

$A$

$=\angle$

$B$

Also,

$\angle AOC=\angle BOD\cdots[\text{vertically opposite angles}]$

So, by

$ASA$ test,

$\Delta$

$AOC$

$\cong\Delta$

$BOD$

And we know that corresponding angles and sides of congruent triangles are equal

So, we can say that

$\angle ACO=\angle BDO$ which are corresponding angles

In the following pairs of right triangles, the measures of some parts are indicated along side. State by the application of RHS congruence condition which are congruent. State the result in symbolic form.

By RHS congruence condition if hypotenuse and one side of one right triangle are equal to hypotenuse and one side of other right triangle then both right triangles are congruent.

In right triangles

$ABC$ and

$ADC$

$BC=DC=4.5\ cm\cdots\cdots[\text{hypotenuse}]$

And

$AC=AC=\text{common side}\cdots\cdots[\text{sides of right triangles}]$

So, by

$RHS$ condition

$\triangle ABC\cong\triangle ADC$

In figure, BD and CE are altitudes of $\Delta$ ABC and BD $=$ CE. State the three pairs of matching parts you have used to answer.(Is $\Delta$ BCD $\cong\Delta$ CBE).

By RHS congruence condition if hypotenuse and one side of one right triangle are equal to hypotenuse and one side of other right triangle then both right triangles are congruent.

It is given that

$BD=CE$ and

$BC$ is the hypotenuse of small triangles

$CEB$ and

$BDC$

So, In right triangles

$BCD$ and

$CBE$

$BC=CB=\text{common side}\cdots\cdots[\text{hypotenuse}]$

And

$BD=CE=\text{given}\cdots\cdots[\text{sides of right triangles}]$

So, by

$RHS$ condition

$\triangle BCD\cong\triangle CBE$

From above, we can see that three pairs of matching part are two equal sides which are used for RHS test and other id right angle of both triangles which need not to be written when using RHS.

In parallelogram $ABCD$ , points $M$ and $N$ have been taken on opposite sides $AB$ and $CD$ respectively such that $AM=CN$ . Show that $AC$ and $MN$ bisect each other.

Consider

$\triangle AMO$ and

$\triangle CNO$

We know that

$AB\parallel CD$

From the figure we know that

$\angle MAO$ and

$\angle NCO$ are alternate angles

It is given that

$AM=CN$

We know that

$\angle AOM$ and

$\angle CON$ are vertically opposite angles

$\angle AOM=\angle CON$

By ASA congruence criterion

$\triangle AMO\cong \triangle CNO$

So we get

$AO=CO$ and

$MO=NO$ (c.p.c.t)

Therefore it is proved that

$AC$ and

$MN$ bisect each other.

In the figure, $ABCD$ is a quadrilateral and $AC$ is one its $AB+BC+CD+DA> 2AC$ diagonals. Prove that

Consider

$\triangle ABC$

We know that

$AB+BC> AC....(i)$

Consider

$\triangle ACD$

We know that

$AD+CD> AC....(ii)$

By adding both the equations we get

$AB+BC+AD+CD> AC+AC$

So we get

$AB+BC+AD+CD> 2AC.....(iii)$

Therefore it is proved that

$AB+BC+AD+CD> 2AC$

In the given figure, $ABCD$ is a quadrilateral in which $AB=AD$ and $BC=DC$ . Prove that
$AC$ bisects $\angle A$ and $\angle C$ .

Consider

$\triangle ABC$ and

$\triangle ADC$

It is given that

$AB=AD$ and

$BC=DC$

$AC$ is common ie.,

$AC=AC$

By SSS congruence criterion

$\triangle ABC\cong \triangle ADC....(i)$

$\angle BAC=\angle DAC$

so we get

$\angle BAE=\angle DAE$

We know that

$AC$ bisects the

$\angle BAD$ ie.,

$\angle C$

In the adjoining figure, $OPQR$ is a square. A circle drawn with centre $O$ cuts the square in $X$ and $Y$ .
Prove that $QX=QY$

Consider

$\triangle OXP$ and

$\triangle OYR$

We know that

$\angle OPX$ and

$\angle ORY$ are right angles

So we get

$\angle OPX=\angle ORY =90^{o}$

We know that

$OX$ and

$OY$ are the radii

$OX=OY$

From the figure, we know that the sides of a square are equal

$OP=OR$

By RHS congruence criterion

$\triangle OXP\cong OYR$

$PX=RY (c.p.c.t.)$

We know that

$PQ=QR$

So we get

$PQ-PX=QR-RY$

$QX=QY$

Therefore, it is proved that

$QX=QY$

If $O$ is a point within a quadrilateral $ABCD$ , show that $OA+OB+OC+OD> AC+BD$

It is given that

$O$ is a point within a quadrilateral

$ABCD$

Consider

$\triangle ACO$

We know that in a triangle the sum of any two sides is greater than the third side

we get

$OA+OC> AC...(i)$

in the same way

Consider

$\triangle BOD$

we get

$OB+OD> BD....(ii)$

by adding both equations we get

$OA+OC+OB+OD> AC+BD$

therefore it is proved that

$OA+OC+OB+OD> AC+BD$

In the given figure, $ABCD$ is a square and $\angle PQR={90}^{o}$ . If $PB=QC=DR$ , prove that $AB+BC+CD+DA> AC+BD$

Consider

$\triangle ABC$

We know that

$AB+BC> AC....(i)$

Consider

$\triangle ACD$

we know that

$AD+CD> AC....(ii)$

By adding both the equations we get

$AB+BC+AD+CD> AC+AC$

so we get

$AB+BC+AD+CD> 2AC.....(iii)$

Consider

$\triangle ABD$ and

$\triangle BDC$

we know that

$AB+DA> BD.....(vii)$

so we get

$BC+CD> BD....(viii)$

by adding(vii) and (viii) we get

$AB+DA+BC+CD> BD+BD$

$AB+DA+BC+CD> 2BD.....(ix)$

By adding equations (ix) and (iii) we get

$AB+DA+BC+CD+AB+BC+AD+CD> 2BD+2AC$

so we get

$2(AB+BC+CD+DA)> 2(BD+AC)$

Dividing by 2

$AB+BC+CD+DA> BD+AC$

Therefore it is proved that

$AB+BC+CD+DA> BD+AC$

In the given figure, $ABCD$ is a quadrilateral in which $AB=AD$ and $BC=DC$ . Prove that
$BE=DE$

Consider

$\triangle ABE$ AND

$\triangle ADE$

It is given that

$AB=AD$

$AE$ is common ie

$AE=AE$

By SAS congruence criterion

$\triangle ABE\cong \angle ADE$

$BE=DE$

In the given figure, $ABCD$ is a square and $\angle PQR={90}^{o}$ . If $PB=QC=DR$ , prove that $AB+BC+CD> DA$

Consider

$\triangle ABC$

We know that

$AB+BC> AC$

Add

$CD$ both sides of the equation

$AB+BC+CD> AC+CD....(iv)$

Consider

$\triangle ACD$

we know that

$AC+CD> DA...(v)$

substituting (v) in (iv) we get

$AB+BC+CD> DA....(vi)$

$\Delta ABC$ is an isosceles triangle with $AB = AC = 13 cm$ . The length of altitude from $A$ on $BC$ is $5 cm$ . Find $BC$ .

In isosceles

$\Delta ABC, \ AB = AC = 13 cm$

Consider

$AL$ is altitude from

$A$ to

$BC$ and

$AL = 5 cm$

Now, in right

$\Delta ALB$

$AB^2 = AL^2 + BL^2$

$(13)^2 = (5)^2 + BL^2$

$169 = 25 + BL^2$

$BL^2 = 169 - 25 = 144$

$BL = 12$

Since

$L$ is midpoint of

$BC$ , then

$BC = 2 \times BC = 2 \times 12 = 24$

$BC$ is

$24 cm$

In the given figure, $ABCD$ is a quadrilateral in which $AB=AD$ and $BC=DC$ . Prove that
$\angle ABC=\angle ADC$

1961122_1715363_ans_3ff60940674e41c6be4b1cff2ab52d24.jpg

Given the base $a$ of a triangle, the opposite angle $A$ and the product $k^2$ of the other sides, solve the triangle and show that there is no such triangle if $a < 2k \sin \dfrac{A}{2}$ .

Let the other two sides of triangle be

$b,c$

Given the base of the triangle is

$a$ and its opposite angle as

$A$

and also

$b c=k^2\implies c=\dfrac{k^2}{b}$

Now

$\cos A=\dfrac{b^2+c^2-a^2}{2 b c}$

$\implies 2 b c\cos A=b^2+c^2-a^2$

$\implies 2 k^2\cos A+a^2=b^2+\bigg(\dfrac{k^2}{b}\bigg)^2$

$\implies b^4-(a^2+2 k^2\cos A)b^2+k^4=0$

Since

$b^2$ is real so

$\text{Discriminant}\ge 0$

$\implies (a^2+2 k^2 \cos A)^2-4 k^4\ge 0$

$\implies (a^2+2 k^2\cos A-2 k^2)(a^2+2 k^2 \cos A+2 k^2)\ge 0$

$(\because (a+b)(a-b)=a^2-b^2)$

$\implies (a^2+2 k^2(2\sin^2 A/2))(a^2+2 k^2(2 \cos^2 A/2))\ge 0$

$(\because \cos 2 A=2\cos^2 A-1=1-2\sin ^2 A)$

$\implies (a^2+4 k^2 \cos^2 A/2)(a^2-4 k^2 \sin^2 A/2)\ge 0$

$\implies a^2-4 k^2\sin^2 \dfrac{A}{2}\ge 0$

$(\because A^2+4 k^2\cos^2 \dfrac{A}{2} \text{ is always positive})$

$\implies \bigg(a-2 k\sin \dfrac{A}{2}\bigg)\bigg(a+2 k \sin \dfrac{A}{2}\bigg)\ge 0$

$a\le -2 k \sin \dfrac{A}{2}$ or

$a\ge 2 k \sin \dfrac{A}{2}$

But

$a$ must be positive which means that

$a\le -2k \sin \dfrac{A}{2}$ is rejected

$\implies a\ge 2 k \sin \dfrac{A}{2}$

$a$ cannot be less than

$2 k \sin \dfrac{A}{2}$

Hence there is no triangle if

$a<2 k \sin \dfrac{A}{2}$

1582768_1734763_ans_c0facb86828a4666806fee0eaff77b26.png

The line segments joining the midpoints $M$ and $N$ are parallel sides $AB$ and $DC$ respectively of a trapezium $ABCD$ is perpendicular to both the sides $AB$ and $DC$ . Prove that $AD = BC$ .

Construct

$AN$ and

$BN$ at the point

$N$

Consider

$\triangle ANM$ and

$\angle BNM$

We know that

$N$ is the midpoint of the line

$AB$

So we get

$AM = BM$

From the figure we know that

$\angle AMN = \angle BMN = 90^{\circ}$

$MN$ is common i.e.

$MN = MN$

$SAS$ congruence criterion

$\triangle ANM \cong \triangle BNM$

$AN = BN (c. p. c. t) \dots(1)$

We know that

$\angle ANM = \angle BNM (c. p. c. t)$

Subtracting

$LHS$ and

$RHS$ by

$90^{\circ}$

$90^{\circ} - \angle ANM = 90^{\circ} - \angle BNM$

So we get

$\angle AND = \angle BNC \dots(2)$

Now, consider

$\triangle AND$ and

$\triangle BNC$

$AN = BN$

$\angle AND = \angle BNC$

We know that

$N$ is the midpoint of the line

$DC$

$DN = CN$

$SAS$ congruence criterion

$\triangle AND \cong \triangle BNC$

$AD = BC (c. p. c. t)$

Therefore, it is proved that

$AD = BC$ .

In the adjoining figure, $X$ and $Y$ are respectively two points on equal sides $AB$ and $AC$ of $\triangle ABC$ such that $AX = AY$ . Prove that $CX = BY$ .

It is given that

$AX = AY$

Considering

$\triangle AXC$ and

$\triangle AYB$

$\angle A$ is common i.e.

$\angle A = \angle A$

We know that the sides of the equilateral triangle are equal so we get

$AC = AB$

$SAS$ congruence criterion

$\triangle AXC \cong \triangle AY$

$BXC = YB (c. p. c. t)$

Therefore, it is proved that

$CX = BY$ .

The angles of a triangle are as $1:2:7$ ; prove that the ratio of the greatest side to the least side is $\sqrt{5}+1:\sqrt{5}-1$ .

1649856_1732973_ans_cf3730df9d5b4af0b7cb82e79a4ceffc.jpg

In $\triangle ABC, AB = AC$ and the bisectors $\angle B$ and $\angle C$ meet at a point $O$ . Prove that $BO = CO$ and the ray $AO$ is the bisector of $\angle A$ .

It is given that

$AB = AC$ and the bisectors

$\angle B$ and

$\angle C$ meet at a point

$O$

Consider

$\triangle BOC$

So we get

$\angle BOC = \dfrac {1}{2} \angle B$ and

$\angle OCB = \dfrac {1}{2} \angle C$

It is given that

$AB = AC$ so we get

$\angle B = \angle C$

So we get

$\angle OBC = \angle OCB$

We know that if the base angles are equal even the sides are equal

So we get

$OB = OC . (1)$

$\angle B$ and

$\angle C$ has the bisectors

$OB$ and

$OC$ so we get

$\angle ABO = \dfrac {1}{2} \angle B$ and

$\angle ACO = \dfrac {1}{2} \angle C$

So we get

$\angle ABO = \angle ACO .. (2)$

Considering

$\triangle ABO$ and

$\triangle ACO$ and equation (1) and (2)

It is given that

$AB = AC$

$SAS$ congruence criterion

$\triangle ABO \cong \triangle ACO$

$\angle BAO = \angle CAO (c. p. c. t)$

Therefore, it is proved that

$BO = CO$ and the ray

$AO$ is the bisector of

$\angle A$ .

In $\triangle ABC, D$ is the midpoint of $BC$ . If $DL \perp AB$ and $DM \perp AC$ such that $DL = DM$ , prove that $AB = AC$ .

It is given that

$D$ is the midpoint of

$BCDL \perp AB$ and

$DM \perp AC$ such that

$DL = DM$

Considering

$\triangle BLD$ and

$\triangle CMD$ as right angled triangle

So we can write it as

$\angle BLD = \angle CMD = 90^{\circ}$

We know that

$BD = CD$ and

$DL = DM$

$RHS$ congruence criterion

$\triangle BLD = \triangle CMD$

$\angle ABD = \angle ACD (c. p. c. t)$

Now, in

$\angle ABC$

$\angle ABD = \angle ACD$

We know that the sides opposite to equal angles are equal so we get

$AB = AC$

Therefore, it is proved that

$AB = AC$ .

A square is inscribed in an isosceles right triangle so that the square and the triangle have one angle common. Show that the vertex of the square opposite the vertex of the common angle bisects the hypotenuse.

$\textbf{Step - 1: Applying properties}$

$\text{Given that }\Delta ABC \text{ is an isosceles right triangle and square }CPQR \text{ inscribed in it.}$

$CPQR \text{ is a square}$

$\therefore CP=PQ=QR=RC$

$\Delta ABC \text{ is an isosceles triangle }$

$\therefore AC=BC$

$\implies AR+RC=CP+BP$

$\implies AR=BP ......(1)(\because RC=CP)$

$\textbf{Step - 2: Applying properties}$

$AR=BP$

$\angle ARQ=\angle QPB =90^\text{o}$

$QR=PQ$

$\therefore \Delta ARQ \approx \Delta QPB$

$\implies AQ=QB$

$\therefore Q\text{ bisects the hypotenuse } AB$

$\textbf{Hence proved.}$

If the angles of a triangle b in A.P. and the lengths of the greatest and least sides be $24$ and $16$ feet respectively, find the length of the third side and the angles, given
$log 2=.30103$ , $log 3=.4771213$ ,
and $L\tan 19^o6'=9.5394287$ , diff. for $1'=4084$ .

Since, angles

$A,B$ and

$C$ of the

$\triangle ABC$ are in

$A.P,$

$\therefore$

$A+C=2B$

And also

$A+B+C=180^o$

$\therefore$

$3B=180^o$

$B=60^o$

Now,

$\tan\dfrac{A-C}{2}=\dfrac{a-c}{a+c}\cot\dfrac{B}{2}=\dfrac{24-16}{24+16}\cot 30^o$

$=\dfrac{8\sqrt{3}}{40}=\dfrac{2\sqrt{3}}{10}$

$L\tan\dfrac{A-C}{2}=\log 2-\log 10+\dfrac{1}{2}\log 3+10$

$L\tan\dfrac{A-C}{2}=0.30103-1+\dfrac{1}{2}\times 0.4771213+10$

$=9+0.30103+0.2385606=9.5395906$

Now

$L\tan\dfrac{A-C}{2}-L\tan 19^o6'=9.5395906+9.5294287=0.0001679$

$\therefore$ Difference for

$60"=0.0004084$

$\therefore$ Difference for

$0.001679=\dfrac{60\times 0.0001679}{0.0004084}=24"$

Hence

$\dfrac{A-C}{2}=19^o6'24"$ ---- ( 1 )

But

$\dfrac{A+C}{2}=90^o-\dfrac{B}{2}=60^o$ ---- ( 2 )

Solving equation ( 1 ) and ( 2 ), we have

$\angle A=79^o6'24",\,\angle C=40^o53'36"$

Again

$\cos B=\dfrac{a^2+c^2-b^2}{2ac}$

$\Rightarrow$

$\cos 60^o=\dfrac{(24)^2+(16)^2-b^2}{2\times 24\times 16}$

$\Rightarrow$

$\dfrac{1}{2}=\dfrac{576+256-b^2}{768}$

$\Rightarrow$

$2\times 832-2b^2=786$

$\Rightarrow$

$b=\sqrt{448}=8\sqrt{7}$

In the given figure, $ABCD$ is a square and $P$ is a point inside it such that $PB = PD$ . Prove that $CPA$ is a straight line.

It is given that

$ABCD$ is a square and

$P$ is a point inside it such that

$PB = PD$

Considering

$\triangle APD$ and

$\triangle APB$

We know that all the sides are equal in a square

So we get

$DA = AB$

$AP$ is common i.e.

$AP = AP$

According to

$SSS$ congruence criterion

$\triangle APD \cong \triangle APB$

We get

$\angle APD = \angle APB (c. p. c. t)\dots (1)$

Considering

$\triangle CPD$ and

$\triangle CPB$

We know that all the sides are equal in a square

So we get

$CD = CB$

$CP$ is common i.e.

$CP = CP$

According to

$SSS$ congruence criterion

$\triangle CPD \cong \triangle CPB$

We get

$\angle CPD = \angle CPB (c. p. c. t)\dots (2)$

By adding both the equation (1) and (2)

$\angle APD + \angle CPD = \angle APB + \angle CPB \dots (3)$

From the figure we know that the angles surrounding the point

$P$ is

$360^{\circ}$

So we get

$\angle APD + \angle CPD + \angle APB + \angle CPB = 360^{\circ}$

By grouping we get

$\angle APB + \angle CPB = 360^{\circ} (\angle APD + \angle CPD) \dots(4)$

Now by substitution of (4) in (3)

$\angle APD + \angle CPD = 360^{\circ} (\angle APD + \angle CPD)$

On further calculation

$2 (\angle APD + \angle CPD) = 360^{\circ}$

By division we get

$\angle APD + \angle CPD = 180^{\circ}$

Therefore, it is proved that

$CPA$ is a straight line.

In the adjoining figure, explain how one can find the breadth of the river without crossing it.

Consider

$AB$ as the breadth of the river. Take a point

$M$ at a distance from

$B$ . Draw a perpendicular from the point

$M$ and name it as

$N$ so that it joins the point

$A$ as a straight line.

Now in

$\triangle ABO$ and

$\triangle NMO$

We know that

$\angle OBA = \angle OMN = 90^{\circ}$

We know that

$O$ is the midpoint of the line

$BM$

So we get

$OB = OM$

From the figure we know that

$\angle BAO$ and

$\angle MON$ are vertically opposite angles

$\angle BAO = \angle MON$

$ASA$ congruence criterion

$\triangle ABO \cong \triangle NMO$

$AB = NM (c. p. c. t)$

Therefore,

$MN$ is the width of the river.

Show that for any triangle with sides a, b and c, $3(ab + bc + ca) < (a + b +c)^2 < 4(bc + ca + ab)$ . when are the first two expressions equal?

We know that for sides

$a, b, c$ of a triangle.

$(a-b)^{2} \geq 0$

$\Rightarrow a^{2}+b^{2} \geq 2 a b$

Similarly,

$b^{2}+c^{2} \geq 2 b c$

$c^{2}+a^{2} \geq 2 c a$

Adding the three inequalities, we get

$2\left(a^{2}+b^{2}+c^{2}\right) \geq 2(a b+b c+c a)$

$\Rightarrow a^{2}+b^{2}+c^{2} \geq a b+b c+c a$

Adding

$2(a b+b c+c a)$ to both sides, we get

$(a+b+c)^{2} \geq 3(a b+b c+c a)$

$3(a b+b c+c a) \leq(a+b+c)^{2}$

Also,

$c<a+b$ (triangle inequality)

$\Rightarrow \quad c^{2}<a c+b c$

Similarly,

$b^{2}<a b+b c$

$a^{2}<a b+c a$

Adding (4),(5) and

$(6),$ we get

$a^{2}+b^{2}+c^{2}<2(a b+b c+c a)$

Adding

$2(a b+b c+c a)$ to both sides, we get

$(a+b+c)^{2}<4(a b+b c+c a)$

Combining (4) and (8), we get

$3(a b+b c+c a) \leq(a+b+c)^{2}<4(a b+b c+c a)$

First two expressions will be equal for

$a=b=c$

Given is a pair of congruent triangles. State the property of congruence and the name of the congruent triangles.

RHS congruence property: Two right triangles are congruent if the hypotenuse and one side of the first triangle are respectively equal to the hypotenuse and one side of the second triangle.

From the figure

$\angle P=\angle N=90 ^{\circ}$ ,

$PQ=NM$ and

$QR=ML$

Comparing the correspondence between sides and angles, We name the congruent triangles as

$\triangle PQR\cong \triangle NML$

Given is a pair of congruent triangles. State the property of congruence and the name of the congruent triangles.

ASA congruence property: Two triangles are congruent if the two angles and the included side of one are respectively equal to the two angles and the included side of the other

From the figure,

$\angle BAC=\angle DAC=50^{\circ}$ ,

$\angle BCA=\angle DCA$ and

$AC$ is common

Comparing the correspondence between sides and angles, We name the congruent triangles as

$\triangle ACB\cong \triangle ACD$

Given is a pair of congruent triangles. State the property of congruence and name of the congruent triangles.

SAS congruence property: Two triangles are congruent if the two sides and the included angle of one are respectively equal to the two sides and the included angle of the other.

From the figure, We have

$\angle C=\angle E$ ,

$CA=ED$ and

$BC=FE$

Comparing the correspondence between sides and angles, We name the congruent triangles as

$\triangle ABC \cong \triangle DFE$

Given is a pair of congruent triangles. State the property of congruence and the name of the congruent triangles.

SSS congruence property: Two triangles are congruent if the three sides of one triangle are respectively equal to the three sides of the other triangle.

From the figure, We have

$XY=RS$ ,

$YZ=ST$ and

$ZX=TR$

Comparing the correspondence between sides, We name the congruent triangles as

$\triangle XYZ\cong \triangle RST$

State the correspondence between the vertices, sides and angles of the following pair of congruent triangles.
$\triangle CAB\cong \triangle QRP$

Two triangles are congruent if pairs of corresponding sides and corresponding angles are equal.

$\therefore \triangle CAB\cong \triangle QRP$ would mean that

Correspondence between vertices:

$C\leftrightarrow Q,A\leftrightarrow R,B\leftrightarrow P$

Correspondence between sides

$CA=QR,AB=RP,BC=PQ$

Correspondence between angles

$\angle C=\angle Q$ ,

$\angle A=\angle R$ ,

$\angle B=\angle P$

State the correspondence between the vertices, sides and angles of the following pair of congruent triangles.
$\triangle XZY\cong \triangle QPR$

Two triangles are congruent if pairs of corresponding sides and corresponding angles are equal

$\therefore\triangle XZY\cong \triangle QPR$ would mean that

Correspondence between vertices

$X\leftrightarrow Q,Z\leftrightarrow P,Y\leftrightarrow R$

Correspondence between sides:

$XZ=QP,ZY=PR,XY=QR$

Correspondence between angles:

$\angle X=\angle Q$ ,

$\angle Z=\angle P$ ,

$\angle Y=\angle R$

Given is a pair of congruent triangles. State the property of congruence and the name of the congruent triangles.

ASA congruence property: Two triangles are congruent if the two angles and the included side of one are respectively equal to the two angles and the included side of the other

From the figure, We have

$\angle E=\angle N$ ,

$\angle F=\angle M$ and

$EF=NM$

Comparing the correspondence between sides and angles, We name the congruent triangles as

$\triangle DEF\cong \triangle PNM$

State the correspondence between the vertices, sides and angles of the following pair of congruent triangles.
$\triangle MPN\cong \triangle SQR$

Two triangles are congruent if pairs of corresponding sides and corresponding angles are equal.

$\therefore \triangle MPN\cong \triangle SQR$ would mean that,

Correspondence between vertices:

$M\leftrightarrow S$ ,

$P\leftrightarrow Q$ ,

$N\leftrightarrow R$

Correspondence between sides:

$MP=SQ,PN=QR,MN=SR$

Correspondence between angles:

$\angle M=\angle S$ ,

$\angle P=\angle Q$ ,

$\angle N=\angle R$

State the correspondence between the vertices, sides and angles of the following pair of congruent triangles.
$\triangle ABC\cong \triangle EFD$

Two triangles are congruent if the corresponding sides and corresponding angles are equal

$\therefore$

$\triangle ABC\cong \triangle EFD$ would mean that

Correspondence between vertices:

$A\leftrightarrow E,B\leftrightarrow F,C\leftrightarrow D$

Correspondence between sides

$AB=EF,BC=FD,CA=DE$

Correspondence between angles:

$\angle A=\angle E$ ,

$\angle B=\angle F$ ,

$\angle C=\angle D$

In the given figure, $PA\bot AB$ , $QB\bot AB$ and $PA=QB$ . Prove that $\triangle OAP=\triangle OBQ$ . Is $OA=OB$ ?

Given:

$PA\bot AB$ ,

$QB\bot AB$ and

$PA=QB$

To prove:

$\triangle OAP=\triangle OBQ$

$OA=OB$ ?

Proof:

$\triangle OAP$ and

$\triangle OBQ$

$PA=QB$ (given)

$\angle POA=\angle QOB$ (vertically opposite angles)

$\angle OAP=\angle OBQ={90}^{o}$

$\therefore\triangle OAP\cong \triangle OBQ$ (From AAS congruence property)

$\Rightarrow OA=OB$ (corresponding parts of the congruent triangles)

In $\triangle ABC, \angle A$ is acute. $BD$ and $CE$ are perpendicular on $AC$ and $AB$ respectively. Prove that $AB\times AE = AC \times AD$ .

Consider the

$\triangle ABC$ ,
So, we have to prove that,

$AB \times AE = AC \times AD$
Now, consider the

$\triangle ADB$ and

$\triangle AEC$ ,

$\angle A = \angle A$ [common angle for both triangles]

$\angle ADB = \angle AEC$ [both angles are equal to

$90^{\circ}$ ]

$\triangle ADB \sim \triangle AEC$
So,

$AB/AC = AD/AE$
By cross multiplication we get,

$AB \times AE = AC \times AD$ .
1700147_1783398_ans_ae39bdfd2d1f4803a53544ae6d207d0a.PNG

In figure, $AD=BC$ and $AD\parallel BC$ . Is $AB=DC$ ? Give reasons in support of your answer.

Given

$AD=BC$ and

$AD\parallel BC$

To prove:

$AB=DC$

Proof:

$\triangle ABC$ and

$\triangle CDA$

$BC=DA$ (given)

$\angle BCA=\angle DAC$ (alternate angles)

$AC=AC$ (common)

$\therefore\triangle ABC\cong \triangle CDA$ [by SAS congruence property]

Hence

$AB=CD$

In the figure, $AB=AD$ and $CB=CD$ . Prove that $\triangle ABC\cong \triangle ADC$ .

Given:

$AB=AD$ and

$CB=CD$

To prove

$\triangle ABC\cong \triangle ADC$

Proof:

$\triangle ABC$ and

$\triangle ADC$

$AB=AD$ (given)

$CB=CD$ (given)

$AC=AC$ (common)

$\therefore$

$\triangle ABC\cong \triangle ADC$ (By SSS congruence property)

In the given figure, triangles $ABC$ and $DCB$ are right-angled at $A$ and $D$ respectively and $AC=DB$ . Prove that $\triangle ABC\cong \triangle DCB$

Given,

$\angle A=\angle D=90^o$

$AC=DB$

To prove,

$\triangle ABC\cong \triangle DCB$

Proof,

$\triangle ABC$ and

$\triangle DCB$

$AC=DB$ (given)

$BC=BC$ (common side)

$\angle CAB=\angle BDC={90}^{o}$

$\therefore\triangle ABC\cong \triangle DCB$ [From the RHS congruence property]

In the adjoining figure, $AB=AC$ and $BD=DC$ . Prove that $\triangle ADB\cong \triangle ADC$ and hence show that (i) $\angle ADB=\angle ADC={90}^{o}$ , (ii) $\angle BAD=\angle CAD$

Given

$AB=AC$ and

$BD=DC$

To prove:

$\triangle \cong \triangle ADC$

Proof:

$\triangle ADB$ and

$\triangle ADC$ , we have

$AB=AC$ (given)

$BD=DC$ (given)

$AD=AD$ (common)

$\therefore$

$\triangle ADB\cong \triangle ADC$ [By SSS congruence property]

(i)

$\angle ADB=\angle ADC$ (corresponding parts of the congruent triangles )

$...(1)$

Now,

$\angle ADB+\angle ADC={180}^{o}$ [

$\because\angle ADB$ and

$\angle ADC$ are on the straight line]

$\Rightarrow \angle ADB+\angle ADB={ 180 }^{ o }$ [from

$(1)$ ]

$\Rightarrow 2\angle ADB={ 180 }^{ o }$

$\Rightarrow \angle ADB=\dfrac{180^{o}}2={90}^{o}$

$\therefore\angle ADB=\angle ADC={90}^{o}$ [from

$(1)$ ]

(ii)

$\angle BAD=\angle CAD$ (

$\because$ corresponding parts of the congruent triangles)

In figure, $PL\bot OA$ and $PM\bot OB$ such that $PL=PM$ . Is $\triangle PLO\cong \triangle PMO$ ?
Give reasons in support of your answer.

Given

$PL\bot OA$ ,

$PM\bot OB$ and

$PL=PM$

To Prove:

$\triangle PLO\cong \triangle PMO$

Proof:

$\triangle PLO$ and

$\triangle PMO$

$\angle PLO=\angle PMO={90}^{o}$

$PO=PO$ (common side)

$PL=PM$ (given)

$\therefore$

$\triangle PLO\cong \triangle PMO$ by the SAS congruence property.

In the given pairs of triangles, applying only the ASA congruence criterion, determine which triangles are congruent. Also, write the congruent triangles in symbolic form.

$\triangle ABD$ and

$\triangle CDB$

$BD=BD$ [Common]

$\angle ABD=\angle CDB=30^{o}$ [Given]

$\angle ADB=\angle CBD=40^{o}$ [Given]

$\therefore \triangle ABD \cong \triangle CDB$ [ASA criterion]

In the given pair of triangles, applying only the ASA congruence criterion, determine which triangles are congruent. Also, write the congruent triangles in symbolic form.

$\triangle XYZ$ and

$\triangle LMN$

$XY=LM=4.8\,cm$ [Given]

$\angle YXZ= \angle MLN=100^{o}$ [Given]

$\angle XYZ=\angle LMN=50^{o}$ [Given]

$\therefore \triangle XYZ \cong \triangle LMN$ [by ASA criterion]

In given pairs of triangles in the figure, applying only the ASA congruence criterion, determine which triangles are congruent. Also, write the congruent triangles in symbolic form.

$\triangle ABC$ ,

$\angle A=55^o$

$\angle B=60^o$ and

$\angle C=180^o-(60^o+55^o)=180^o-115^o=65^o$

$\triangle PQR$

$\angle Q=55^o$

$\angle P=50^o$

Since none of the angles in

$\triangle ABC$ is equal to

$\angle P$ .

$\therefore \triangle ABC$ and

$\triangle PQR$ are not congruent.

In the given pairs of triangles, applying only the ASA congruence criterion, determine which triangles are congruent. Also, write the congruent triangles in symbolic form.

$\triangle PON$ and

$\triangle MNO$

$ON=NO$ [Common]

$\angle NOP=\angle ONM =90^{o}$ [Given]

$\angle ONP=\angle NOM=50^{o}$ [Given]

$\therefore \triangle PON \cong \triangle MNO$ [by ASA criterion]

In Fig., $\Delta PQR \cong \Delta$ ______

$\triangle PQR$ and

$\triangle RSP$

$QR = SP = 4.1 \,cm$ [Given]

$\angle PQR = \angle RSP = 45^o$ [Given]

$PR = RP$ [Common]

$\therefore \triangle PQR \cong \triangle RSP$ [ by SAS criterion]

In the given pairs of triangles, applying only the ASA congruence criterion, determine which triangles are congruent. Also, write the congruent triangles in symbolic form.

$\triangle AOD$ and

$\triangle BOC$

$\angle AOD=\angle BOC$ [vertically opposite angles]

$OD=OC$ [Given]

$\angle ADO=\angle BCO$ [Given]

$\therefore \triangle AOD \cong \triangle BOC$ [by ASA criterion]

If $\Delta PQR$ and $\Delta XYZ$ are congruent under the correspondence $QPR \leftrightarrow XYZ$ , then,
(i) $\angle R =$ (ii) $QR =$
(iii) $\angle P =$ (iv) $QP =$
(v) $\angle Q =$ (vi) $RP =$

Given

$\triangle PQR\cong \triangle XYZ$ under the correspondence

$QPR\leftrightarrow XYZ$

Therefore,

(i)

$\angle R=\angle Z$

(ii)

$QR=YZ$

(iii)

$\angle P=\angle X$

(iv)

$QP =YZ$

(v)

$\angle Q=\angle Y$

(vi)

$RP=ZX$

In the given pair of triangles, using only the RHS congruence criterion, determine which pairs of triangles are congruent. In case of congruence. write the result in symbolic form

$\triangle XYZ$ and

$\triangle UZY$

$\angle Y=\angle Z=90^{o}$ [Given]

$XZ=UY$ [Hyptonenuse]

$YZ=ZY$ [Common]

$\therefore \triangle XYZ \cong \triangle UZY$ [by RHS criterion]

Enter 1 if it is true else enter 0.
In Fig., $\triangle PQR \cong \triangle$ $XYZ$

$\triangle PQR$ and

$\triangle XYZ$ ,

$PQ = XZ = 3.5 \,cm$ [Given]

$\angle PQR = \angle XYZ = 45^o$ [Given]

$QR = ZY = 5 \,cm$ [Given]

$\therefore \triangle PQR \cong \triangle XYZ$ [ by SAS Criterion]

In Fig., $\Delta$ ______ $\cong PQR$

$\triangle PQR$ and

$\triangle DRQ$

$\angle PQR = \angle DRQ = 40^o$ [Given]

$\angle PQR = \angle DRQ = 30^o + 40^o = 70^o$ [Given]

$QR=QR$ [Common]

$\therefore \Delta PQR \cong \Delta DRQ$ [ by ASA criterion]

In Fig., $AB = AD$ and $\angle BAC = \angle DAC$ .
Then,
(i) $\Delta$ $\cong \Delta ABC$
(ii) $BC =$
(iii) $\angle BCA =$
(iv) Line segment $AC$ bisects and __.

(i) In

$\triangle ABC$ and

$\triangle ADC$

$AC = AC$ [Common]

$\angle BAC = \angle DAC$ [Given]

$AB = AD$ [Given]

$\therefore \triangle ABC = \triangle ADC$ [by SAS criterion]

(ii) We have,

$\triangle ABC = \triangle ADC$

$\Rightarrow BC = DC$ [by C.P.C.T.]

(iii) and,

$\angle BCA = \angle DCA$ [by C.P.C.T.]

(iv) Given that

$\angle BAD=\angle BCD$

Hence, Line segment

$AC$ bisects

$\angle BAD$ and

$\angle BCD$

In the given pair of triangles, using only the RHS congruence criterion, determine which pairs of triangles are congruent. In case of congruence. write the result in symbolic form:

$\triangle ADB$ and

$\triangle CDB$

$AD=AD$ [Common]

$\angle ADB= \angle ADAC=90^{o}$ [Given]

$AB=AC$ [Hypotenuse]

$\therefore \triangle ADB \cong \triangle ADC$ [ by RHS criterion]

In the given pair of triangles, applying only the ASA congruence criterion, determine which triangles are congruent. Also, write the congruent triangles in symbolic form.

$\triangle ABC$ and

$\triangle DFE$

$\angle B=\angle F$ [Given]

$\angle C=\angle E$ [Given]

$BC=FE$ [Given]

$\therefore \triangle ABC \cong \triangle DFE$ [by ASA criterion]

Without drawing the triangles write all size pairs of equal measures in each of the following pairs of congruent triangles.
$\triangle XYZ \cong \triangle MLN$

$\triangle XYZ \cong \triangle MLN$

$XY=ML, YZ=LN, XZ=MN$

$\angle XYZ=\angle MLN, \angle XZY=\angle MNL, \angle ZXY=\angle NML$

In the following, pairs of triangles of Fig. the lengths of the sides are indicated along the sides. By applying SSS congruence criterion, determine which triangles are congruent. If congruent, write the results in symbolic form.

$\triangle LMN$ and

$\triangle GHI$ ,

Given,

$LM = GH = 4.5cm$

$MN = HI = 6cm$

$MN = GI = 5cm$

$\therefore$

$\triangle LMN \cong \triangle GHI$ ....SSS criterion

In the following, pairs of triangles of FIg. the lengths of the sides are indicated along the sides. By applying SSS congruence criterion, determine which triangles are congruent. If congruent, write the results in symbolic form.

In triangle

$ABC$ and

$NLM$ ,

Given,

$NL = AB = 5cm$

$BC = LM = 6cm$

$AC = NM = 4cm$

$\therefore$

$\triangle ABC \cong \triangle NLM$ ...SSS criterion

In the given pairs of triangles, using only the RHS congruence criterion, determine which pairs of triangles are congruent. In case of congruence. write the result in symbolic form:

$\triangle ABC$

Using Pythagoras Theorem

$(AC)^{2}=(AB)^{2}+(BC)^{2}$

$\Rightarrow (AC)^2=6^{2}+8^{2}$

$\Rightarrow (AC)^2=36+64$

$\Rightarrow AC=\sqrt{100}$

$\Rightarrow AC=10\,cm$

$\triangle EDC,$

$DC=BD-BC$

$\Rightarrow (14-8)cm$

$\Rightarrow DC=16\,cm$

$\Rightarrow CE=10\,cm$

Now, in

$\triangle ABC$ and

$\triangle CDE$

$\angle B=\angle D=90^{o}$ [Given]

$AB=CD=6\,cm$

$AC=CE=10\,cm$ [Hypotenuse]

$\therefore \triangle ABC \cong \triangle CDE$ [by RHS criterion]

In Fig. which phase of triangles are congruent by SAS congruence criterion (condition)? If congruent, write the congruence of the two triangles in symbolic form.

In triangle

$ABE$ and

$CBD$ ,

$\angle EAB = \angle DCB = 50^{\circ}$

$CD = AE = 5\ cm$

$CB = AB = 5.2\ cm$

$\therefore$

$\triangle CBD \cong \triangle ABE$ ....SAS criterion

In the given pairs of triangles, using only the RHS congruence criterion, determine which pairs of triangles are congruent. In case of congruence. write the result in symbolic form:

$\triangle AOD$ and

$\triangle BOC$

$LO=LO$ [Common]

$LM=LN=8\,cm$ [Hypotenuse]

$\angle LOM=\angle LON=90^{o}$ [Given]

$\therefore \triangle LOM \cong \triangle LON$ [by RHS criterion]

State which of the following pairs of triangles are congruent. If yes, write them in symbolic form (you may draw a rough figure).
$\triangle PQR : PQ = 3.5cm, QR = 4.0cm, \angle Q = 60^{o}$
$\triangle STU : ST = 3.5cm, TU = 4cm, \angle T = 60^{o}$

In triangle

$PQR$ and

$STU$ ,

Given,

$PQ = ST = 3.5cm$

$\triangle PQR = \triangle STU = 60^{o}$

$\Rightarrow$

$OR = TU = 4cm$

$\therefore$

$\triangle STU \cong \triangle PQR$ ... SAS criterion

Without drawing the triangles write all size pairs of equal measures in each of the following pairs of congruent triangles.
$\triangle YZX \cong \triangle PQR$

$\triangle YZX \cong \triangle POR$

$YZ=PQ, ZX=OR, YX=PR$

$\angle YZX=\angle POR, \angle YXZ=\angle PRO, \angle ZYZ=\angle RPQ$

In the given pair of triangles, using only RHS congruence criterion, determine which pairs of triangles are congruent. In case of congruence. write the result in symbolic form:

$\triangle ACE\ \&\ \triangle BDE$

$\angle ACE=\angle BDE=90^o$ [Given]

$AB=BE$ [Hypotenuse]

$CB=BD$ [Given]

$\therefore \triangle ACE \cong \triangle BDE$ [by RHS criterion]

In the given pairs of triangles, using only the RHS congruence criterion, determine which pairs of triangles are congruent. In case of congruence. write the result in symbolic form:

$\triangle XNY$ and

$\triangle XUY$

$XY=XY$ [Common]

$XN\not=XU$

Hence, Triangle

$XYZ$ and

$YXU$ are not congruent with any criterion.

Observe, Figure and state the three pairs of equal parts in triangles ABC and DBC.
$(i) Is \triangle ABC \cong \triangle DCB? Why?$
$(ii)Is AB=DC?Why?$
$(iii)Is AC=DB?Why?$

(i)In triangle

$DCB$ and

$ABC$ ,

Given

$\Rightarrow \angle ABC=\angle DCB=40^{o}+30^{o}$

$\Rightarrow 70^{o}$

$\angle ACB=\angle DBC=30^{o}$

$CB=BC$

$\therefore \triangle ABC \cong \triangle DCB$ .....ASA criterion

(ii) By using the part(1)of a solution, we get

$AB=DC$ ...By C.P.C.T

(iii)By using the part(1) of a solution we get

$AC=DB$ .....By C.P.C.T

In Figure, state the three pairs of equal parts in $\triangle ABC$ and $\triangle EOD$ . Is $\triangle ABC \cong \triangle EOD$ ?Why?

In triangle

$ABC$ and

$EOD$

$AE=EO$ ....given

$\angle ABC=90^{o}=\angle EOD$ ...given

$ED=AC$ ...hypotenuse

$\therefore \triangle ABC \cong \triangle EOD$ ...RHS criterion

In Figure $QS \bot PR$ , $RT \bot PQ$ and $QS=RT$ .
(i) Is $\triangle QSR \cong \triangle RTQ$ ? Give Reasons
(ii) Is ∠ PQR = ∠ PRQ? Give reasons.

(i) In triangle RTO and QSR

Given,

$\Rightarrow \angle RTO=\angle OSR=90^{o}$ (

$QS \bot PR$ and

$RT \bot PQ$ )

$\Rightarrow QS=RT$

$\Rightarrow QR=RO$ ....Hypotenuse

$\therefore \triangle RTQ \cong \triangle QSR$ ....RHS criterion

(ii) By using part(1) of a solution, we get

$\angle TQR=\angle SRQ$ ....By C.P.C.T

$\Rightarrow \angle PQR=\angle PRQ$

In above figure, $\angle 1=\angle 2$ and $\angle 3 =\angle 4$ .
(i) Is $\triangle ADC \cong \triangle ABC$ ? Why?
(ii)Show that $AD=AB$ and $CD=CB$ .

(i) In triangle

$ABC$ and

$ADC$ ,
Given,

$\angle 1=\angle 2$ {Given}

$AC=AC$ {Common}

$\angle 3=\angle 4$ {Given}

$\therefore \triangle ABC \cong \triangle ADC$ .....ASA criterion for congruency

(ii) By using the part (a) of a solution, we get

$\begin{bmatrix}AB=&AD\\CB=&CD\end{bmatrix}$ {By C.P.C.T}

In the figure shown above, $DE=IH, EG=FI$ and $\angle E=\angle I$ . Is $\triangle DEF \cong \triangle HIG?$ If yes, by which congruence criterion?

$\triangle HIG$ and

$\triangle DEF$ ,

Given,

$HI=DE\quad \quad ....(1)$

$\angle E=\angle I\quad \quad ....(2)$

$FI=EG$

$\therefore \ IF+GF=EG+GF$

$\Rightarrow IG=EF\quad \quad ....(3)$

According to the SAS congruence criterion,

$\triangle HIG \cong \triangle DEF$

$\quad \quad [(1),\ (2)\ and \ (3)]$

$\therefore \ \triangle HIG \cong \triangle DEF$ by the SAS congruence criterion.

$AD$ is a median of the triangle $ABC$ . Is it true that $AB+BC+CA>2AD$ ? Give reason for your answer.

$\triangle \ ABD$ , we have

$AB+BD>AD....(1)$
[

$\therefore$ Sum of the lengths of any two sides of a triangle must be greater than the third side]
Now,

$\triangle \ ADC$ , we have

$AC+CD>AD....(2)$
[

$\therefore$ Sum of the lengths of any two sides of a triangle must be greater than the third side]
Adding

$(1)$ and

$(2)$ , we get

$AB+BD+CD+AC>2AD$

$\implies \ AB+BC+CA>2AD$ [

$\therefore BD=CD$ as

$AD$ is median of

$\triangle \ ABC$ ]

1809262_1794715_ans_809ea4614a8d47c8929450f353783b6b.png

State which of the following pairs of triangles are congruent. If yes, write them in symbolic form (you may draw a rough figure).
$\triangle ABC : AB = 4.8cm, \angle A = 90^{o}, AC = 6.8cm$
$\triangle XYZ : YZ = 6.8cm, \angle X = 90^{o}, ZX = 4.8cm$

The given triangles

$ABC$ and

$XYZ$ are not congruent by any criterion of congruence.
1670010_1793430_ans_3fe9bbc8e6f143fc8d061926ad8c4d31.png

In Fig. $PQ = PS$ and $\angle 1 = \angle 2$ .
(i) Is $\triangle PQR \cong \triangle PSR$ ? Give Reasons.
(ii) Is $QR = SR$ ? Give reasons.

(i) In triangle

$PSR$ and

$PQR$ ,

Given,

$PS = PQ$

$\angle 1 = \angle 2$

$PR = PR$

$\therefore$

$\triangle PSR \cong \triangle POR$ .... SAS criterion

(ii) Yes,

$SR = QR$ .... By C.P.C.T

If two sides and an angle of one triangle are equal to two sides and an angle of another triangle, then the two triangles must be congruent. Is the statement true? Why?

No, it is not true statement as the angles should be included angle of there two given sides.

In $\Delta$ PQR , PD QR such that D lies on QR . If $PQ = a , PR = b , QD = c$ and $DR = d$ , prove that $(a + b) (a - b) = (c + d)(c - d)$ .

Given : In

$\Delta PQR, PD \bot QR$ So

$\angle 1 = \angle 2$

$PQ = a, PR = b, QD = c$ and

$DR = d$ .

To prove :

$(a +b)(a-b) = (c+d)(c-d)$

Proof: In right angled

$\angle PDQ$ ,

$PD^2 = PQ^2 - QD^2$ [By Pythagoras theorem]

$\Rightarrow PD^2 = a^2-c^2$ (i)

Similarly, in right angled

$\Delta PDR$ ,

PD^2 = PR^2-DR^2$$ [By Pythagoras theorem]

$\Rightarrow PD^2= b^2 - c^2$ (ii)

From (i) and (ii), we have

$a^2 - c^2 = b^2 - d^2$

$\Rightarrow a^2 - b^2 = c^2 - d^2$

$\Rightarrow (a-b)(a+b) = (c-d)(c+d)$

Hence, proved.

1765336_1796721_ans_9cbda187aa6e4cd5a5bfac23c5c8cad2.png

In the given figure, $AB=AC, P$ and $Q$ are points on $BA$ and $CA$ respectively such that $AP=AQ$ . Prove that
$\triangle APC\cong \triangle AQB$

In the given figure

$AB=AC$

$P$ and

$Q$ are point on

$BA$ and

$CA$ produced respectively such that

$AP=AQ$
Now we have to prove

$\triangle APC \cong \triangle AQB$

$AB=AC$

$AQ+AP$

and

$\angle BAQ=\angle CAP$ (opposite angle)

Hence

$\triangle APC \cong \triangle AQB$

By using corresponding parts of congruent triangle concept we have

$CP=BQ$

$\angle ACP =\angle ABQ$

In the above figure, $BA \perp AC, \ ED \perp FD$ such that $BA=DE$ and $BF=EC$ . Show that $\triangle \ ABC \cong \triangle \ DEF$ .

We have

$BF=EC$

$\therefore$

$BF+FC=CE+FC \implies BC=EF$
In

$\triangle \ ABC, \ \angle A =90^\circ$ and in

$\triangle \ DEF, \ \angle D=90^\circ$ .

$\therefore \triangle \ ABC$ and

$\triangle \ DEF$ are right triangles.
Now, in right triangles

$ABC$ and

$DEF$ , we have

$BA=DE$ [Given]

$\angle A= \angle D \ (Each \ 90^\circ)$
And

$BC=EF$ [Proved above]

$\therefore$

$\triangle \ ABC \cong \triangle \ DEF$ [By RHS congruence rule]

The longest side of a triangle is $3$ times the shortest side and the third side is $2\ cm$ shorter than the longest side. If the perimeter of the triangle is at least $61\ cm$ , find the minimum length of the shortest side.

Let the length of the shortest side of the triangle be

$x\ cm .$

Then, length of the longest side

$=3x\ cm$

Length of the third side

$(3x−2)\ cm$

Since the perimeter of the triangle is at least

$61 cm$

$x\ cm+3x\ cm+(3x−2)\ cm≥61\ cm$

$⇒7x−2≥61$

$⇒7x≥61+2$

$⇒7x≥63$

$⇒\dfrac{7x}7≥\dfrac{63}7$

$⇒x≥9$

Thus, the minimum length of the shortest side is

$9\ cm$ .

We know that in a triangle, the sum of lengths of any two sides is greater than the length of the third side. Is the sum of any angles of a triangle also greater than the third angle? If no, draw a rough sketch to show such a case.

In a triangle, the sum of any two sides must be
greater than its third side but in case of its angles.
It is not necessary that the sum of any two angles be more than its third angle. Such as in this triangle sum of any two angles is less than its third angle.
Such as

$30^\circ + 20^\circ = 50^\circ < 130^\circ$
1732918_1802100_ans_1a02f6179939448cbf16fe3de82478cf.PNG

In the given figure, $ABCD$ is a quadrilateral in which $AD=BC$ and $\angle DAB=CBA$ . Prove that
$\triangle ABD \cong \triangle BAC$

The figure

$ABCD$ is a quadrilateral
In which

$AD=BC$

$\angle DAB =\angle CBA$
To prove:
(i)

$\triangle ABD =\triangle BAC$
(ii)

$\angle ABD =\angle BAC$
Proof: In

$\triangle ABD$ and

$\triangle ABC$

$AB=AB$ (common)

$\angle DAB =\angle CBA$ (Given)

$AD=BC$
(i)

$\triangle ABD\cong \triangle ABC$ (

$SAS$ axiom)
(ii)

$BD=AC$
(iii)

$\angle ABD=\angle BAC$
1714589_1810995_ans_d98992037ed14415a095344762a9eda1.PNG

In the above figure, $AD$ is the bisector of $\angle BAC$ . Prove that $AB>BD$ .

Since exterior angle of a triangle is greater than either of the interior opposite angles, therefore, in

$\triangle \ ACD$ ,
Ext

$\angle ADB> \angle DAC \implies \ ADB> \angle BAD$
[

$\therefore \ AD$ is the bisector of

$\triangle BAC$ , so

$\angle DAC = \angle BAD$ ]
Now, in

$\triangle ABD$ , we have

$\angle ADB> \angle BAD$
Hence,

$AB>BD$ . [

$\therefore$ In a triangle, side opposite to greater angle is longer]

In the adjoining figure, $AB=CD, CE=BF$ and $\angle ACE =\angle DBF$ . Prove that
$\triangle ACE\cong \triangle DBF$

$\triangle ACE$ and

$\triangle DBF,$

And given:

$AB=CD$

Add

$BC$ to both sides

we get

$AB+BC=BC+CD$

$\Rightarrow AC=BD$

$...(1)$

Now,

$CE=BF$ [Given]

$\Rightarrow \angle ACE=\angle DBF$ [Given]

and

$AC=BD$ [From

$(1)$ ]

By SAS congruence rule,

$\triangle ACE \cong \triangle DBF$

In the adjoining figure, $ABCD$ is a square $P,Q$ and $R$ are points on the sides $AB, BC$ and $CD$ respectively such that $AP=BQ=CR$ and $\angle PQR =90^o$ . Prove that
$PQ=QR$

Given: in the figure,

$ABCD$ is a square

$P,Q$ and

$R$ are the Points on the sides

$AB,$

$BC$ and

$CD$ respectively such that

$AP=BQ=CR,\angle PQR=90^{0}$

To prove: (a)

$\triangle PBQ=\triangle QCR$

(b)

$PQ=QR$

(c)

$\angle PQR=45^{0}$

Proof:

$AB=BC=CD$ (sides of square)

And

$AP=BQ=CR$ (Given)

Subtracting, we get

$AB-AP=BC-BQ=CD-CR$

$(PR=QC=RD)$

Now in

$\triangle PBQ$ and

$\angle QCR$

$PB=QC$ (Proved)

$BQ=CR$ (Given)

$\angle B=\angle C$ (Each

$90^{0}$ )

$\triangle PBQ \cong \triangle QCR$

$PQ=QR$

But

$\angle PQR=90^{0}$

$\angle RPQ=\angle PQR$

(Angles opposite to equal angles)

But

$\angle RPQ+\angle PRQ=90^{0}$

$\angle RPQ=\angle PRQ=90^{0}$

$\angle RPQ=\angle PRQ=90^{0}/2=45^{0}$

State, whether the pairs of triangles given in the following figures are congruent or not:

In the right triangles, one side and hypotenuse of one triangle are equal to the corresponding side and hypotenuse of the other.
Therefore, the given triangles are congruent. (RHS Axiom)

In the figure given $ABCD$ is a parallelogram. $E$ and $F$ are mid-point of the sides $AB$ and $CO$ respectively. The straight lines $AF$ and $BF$ meet the straight lines $ED$ and $EC$ in points $G$ and $H$ respectively. Prove that $\triangle HEB\cong \triangle HCF$

Given that,

$ABCD$ is a parallelogram

$E$ and

$F$ are the mid points of sides

$AB$ and

$CD$

To prove:

$\triangle HEB \cong\triangle HCF$

Proof:

in parallelogram

$ABCD$ ,

$\angle CEB =\angle ECF\quad$ [ Alternate angles ]

$\Rightarrow \angle HEB =\angle HCF \qquad...(1)$

$\angle EBF =\angle CFB\quad$ [ Alternate angles ]

$\Rightarrow \angle EBH =\angle CFM \qquad... (2)$

Also,

$E$ and

$F$ are mid point of

$AB$ and

$CD$

$BE=\dfrac12 AB\qquad ...(3)$

$CF =\dfrac12 CD\qquad...(4)$

We know that

$ABCD$ is a parallelogram

$\Rightarrow AB=CD$

$\Rightarrow \dfrac 12 AB =\dfrac 12 CD$

Using equation (3) and (4),

$\Rightarrow BE=CF\qquad...(5)$

Therefore In

$\triangle HEB$ and

$\triangle HCF$ ,

$\angle HEB =\angle FCH\quad$ [ from eq (1)]

$\angle EBH =\angle CFH\quad$ [ from eq (2)]

$BE=CF\quad$ [ from eq (5)]

Hence,

$\triangle HEB \cong \triangle FCH\qquad$ [

$ASA$ congruency criterion ]

In the adjoining figure, $ABCD$ is a square $P,Q$ and $R$ are points on the sides $AB, BC$ and $CD$ respectively such that $AP=BQ=CR$ and $\angle PQR =90^o$ . Prove that
$\angle PRQ = 45^o$

Given: in the figure,

$ABCD$ is a square

$P,Q$ and

$R$ are the Points on the sides

$AB,$

$BC$ and

$CD$ respectively such that

$AP=BQ=CR,\angle PQR=90^{0}$

To prove: (a)

$\triangle PBQ=\triangle QCR$

(b)

$PQ=QR$

(c)

$\angle PQR=45^{0}$

Proof:

$AB=BC=CD$ (sides of square)

And

$AP=BQ=CR$ (Given)

Subtracting, we get

$AB-AP=BC-BQ=CD-CR$

$(PR=QC=RD)$

Now in

$\triangle PBQ$ and

$\angle QCR$

$PB=QC$ (Proved)

$BQ=CR$ (Given)

$\angle B=\angle C$ (Each

$90^{0}$ )

$\triangle PBQ \cong \triangle QCR$

$PQ=QR$

But

$\angle PQR=90^{0}$

$\angle RPQ=\angle PQR$

(Angles opposite to equal angles)

But

$\angle RPQ+\angle PRQ=90^{0}$

$\angle RPQ=\angle PRQ=90^{0}$

$\angle RPQ=\angle PRQ=90^{0}/2=45^{0}$

In the given figure, $PQ || BA$ and $RS\ CA$ . If $BP=RC$ , prove that :
$RS=CQ$

$PQ || BA, RS || CA$

$BP = RC$

To prove:

(i)

$\triangle BSR \cong \triangle PQC$

(ii)

$RS=CQ$

Proof:

$BP =RC$

$BC -RC=BC-BP$

$BR=PC$

Now, in

$\triangle BSR$ and

$\triangle PQC$

$\angle B =\angle P$ (Corresponding angles)

$\angle R =\angle C$ (Corresponding angles)

$BR =PC$ (Proved)

$\triangle BSR \cong \triangle PQC$

$BS=PQ$

$RS=CQ$

1714630_1811031_ans_9c94b388d93f46f2a864213d2cc8d915.PNG

In the adjoining figure, $ABCD$ is a square, $P,Q$ and $R$ are points on the sides $AB, BC$ and $CD$ respectively such that $AP=BQ=CR$ and $\angle PQR =90^o$ . Prove that $\triangle PBQ \cong \triangle QCR$ .

Given: in the figure,

$ABCD$ is a square

$P,Q$ and

$R$ are the Points on the sides

$AB,$

$BC$ and

$CD$ respectively such that

$AP=BQ=CR,\angle PQR=90^{0}$

To prove: (a)

$\triangle PBQ \cong \triangle QCR$

Proof:

$AB=BC=CD$ ....... (1) (sides of square)

And

$AP=BQ=CR$ . ..... (2) (Given)

Subtracting the above equation, we get

$AB-AP=BC-BQ=CD-CR$

$(PB=QC=RD)$

Now in

$\triangle PBQ$ and

$\triangle QCR$

$PB=QC$ (Proved)

$BQ=CR$ (Given)

$\angle B=\angle C$ (Each

$90^{0}$ )

$\triangle PBQ \cong \triangle QCR$ by SAS rule

In triangle $ABC$ and $DEF, \angle A= \angle D, \angle B=\angle E$ and $AB =EF$ . Will the two triangle be congruent? Give reasons for your answer.

$\triangle ABC$ and

$\triangle DEF$

$\angle A= \angle D$

$\angle B=\angle E$

$AB =EF$
In

$\triangle ABC,$ two angles and include side is
Given but in

$\triangle DEF,$ corresponding angles are
Equal but Side is not include of there angle.
Triangles cannot be congruent.

State, whether the pairs of triangles given in the following figures are congruent or not:

In the given figure,
the corresponding three sides are equal.
Therefore, the given triangles are congruent. (SSS Axiom)

State, whether the pairs of triangles given in the following figures are congruent or not:
$\Delta ABC$ in which $AB = 2\,cm , BC = 3.5 \,cm$ and $\angle C = 80^{\circ}$ and $\Delta DEF$ in which $DE = 2\,cm , DF = 3.5\,cm$ and $\angle D = 80^{\circ}$

$\Delta ABC$ ,
It is given that

$AB = 2\,cm , BC = 3.5 \,cm . \angle C = 80^{\circ}$
In

$\Delta DEF$ ,
It is given that

$DE = 2\,cm , DF = 3.5\,cm$ and

$\angle D = 80^{\circ}$
We get to know that two corresponding sides are equal but the included angles are not equal.
therefore , the triangles need not be congruent.
1746062_1831066_ans_5e1c8f342f284514884bf8e7d47e8d0d.png

1746062_1831066_ans_5e1c8f342f284514884bf8e7d47e8d0d.png

$\mathrm{ABCD}$ is a rectangle and $\mathrm{P}$ is mid-point of $\mathrm{AB}$ . DP is produced to meet $\mathrm{CB}$ at $\mathrm{Q}$ . Prove that area of rectangle $\mathrm{ABCD}=$ area of $\Delta \mathrm{DQC}$ .

Given:

$\mathrm{ABCD}$ is a rectangle and

$\mathrm{P}$ is mid-point of

$\mathrm{AB}$ .
DP is produced to meet

$\mathrm{CB}$ at

$\mathrm{Q}$ .
To prove:
area of rectangle

$\mathrm{ABCD}=$ area of

$\Delta \mathrm{DQC}$
Proof :
In

$\triangle \mathrm{APD}$ and

$\triangle \mathrm{BQP},$

$\mathrm{AP}=\mathrm{BP}[\text { since, } \mathrm{D} \text { is the mid-point of } \mathrm{AB}]$

$\angle \mathrm{DAP}=\angle \mathrm{QBP}\left[\text { each angle is } 90^{\circ}\right]$

$\angle \mathrm{APD}=\angle \mathrm{BPQ}[\text { vertically opposite angles }]$
So,

$\Delta \mathrm{APD} \cong \Delta \mathrm{BQP}$ [By using ASA postulate]
ar

$(\Delta \mathrm{APD})=\operatorname{ar}(\Delta \mathrm{BQP})$
Now,

$\operatorname{ar} A B C D=\operatorname{ar}(\Delta A P D)+\operatorname{ar} P B C D$

$\begin{array}{l}=\operatorname{ar}(\Delta \mathrm{BQP})+\operatorname{ar} \mathrm{PBCD} \\=\operatorname{ar}(\Delta \mathrm{DQC})\end{array}$
Hence proved.
1725610_1813080_ans_c3ce25eba6ac4f9e94a6c5bf40a3f255.png

1725610_1813080_ans_c3ce25eba6ac4f9e94a6c5bf40a3f255.png

Prove that
$\angle ADB = \angle ADC$

From the figure

$AD\perp BC$
then,

$\angle ADB=90^\circ\,\&\,\angle ADC=90^\circ$

$\therefore\,\angle ADB=\angle ADC$
1746348_1831099_ans_9d9cf6f0f31b4321bc4d92ee7567579c.png

Prove that
$\Delta ABD \cong \Delta ACD$

From the figure

$BD = CD$
In

$\Delta ABD$ and

$\Delta ACD$

$BD = CD$

$AD = AD$ is common

$\angle ADB=\angle ADC=90^o$

$\Delta ABD \cong\Delta ACD$ (SAS Axiom)
Therefore , it is proved.
1746124_1831092_ans_18c2e61703de4db0aff04f36b600ead5.png

From the given figure, prove that:
$AB = ED$

$\Delta ACB$ and

$\Delta ECD$ ,

$AC = CE$ {Given}

$BC = CD$ {Given}

$\angle ACB = \angle DCE$ (vertically opposite angles)

Thus,

$\Delta ACB \cong \Delta ECD$ (by SAS rule)
Therefore,

$AB = ED$ (by c.p.c.t)

In the given figure , prove that
$\Delta ACB \cong \Delta ECD$

$\Delta ACB$ and

$\Delta ECD$

It is given that

$AC = CE$
and

$BC = CD$

$\angle ACB = \angle DCE$ (are vertically opposite angles)

Hence ,

$\Delta ACB \cong \Delta ECD$ (SAS Axiom)

Prove that
$\angle A = \angle C$

$\Delta ABD$ and

$\Delta BDC$
It is given that

$AB = DC$
and

$BC = AD$

$BD = BD$ is common
Here ,

$\Delta ABD \cong \Delta BDC$ (SSS Axiom)
Here

$\angle A = \angle C$ (c.p.c.t)
Therefore , it is proved

Prove that
$\Delta ABD \cong \Delta BDC$

$\Delta ABD$ and

$\Delta BDC$
It is given that

$AB = DC$
and

$BC = AD$

$BD = BD$ is common
Here ,

$\Delta ABD \cong \Delta DBC$ (SSS Axiom)

In the given figure , prove that
$\Delta ABD \cong \Delta ACD$

$\Delta ABD$ an

$\Delta ACD$

$AD = AD$ is common
It is given that

$AB = AC$
and

$BD = DC$
Here

$\Delta ABD \cong \Delta ACD$ (SSS Axiom)
Therefore , it is proved

Prove that
$\angle ADB = 90^{\circ}$

From the figure

$AD = AC$ and

$BD = CD$
In

$\Delta ABD$ and

$\Delta ACD$

$AD = AD$ is common
We know that

$\angle ADB + \angle ADC = 180^{\circ}$ is a linear pair
Here

$\angle ADB = \angle ADC$
So we get

$\angle ADB = \dfrac{180^{\circ}}{2}$

$\angle ADB = 90^{\circ}$

Prove that :
(i) $\Delta ABC \cong \Delta ADC$
(ii) $\angle B = \angle D$
(iii) $AC$ bisects angle $DCB$ .

In the given figure

$AB = AD$ and

$CB = CD$
In

$\Delta ABC$ and

$\Delta ADC$

$AC = AC$ is common
It is given that

$AB = AD$ and

$CB = CD$

Here

$\angle ABC \cong \Delta ADC$ (SSS Axiom)

$\angle B = \angle D$ (c.p.c.t)
So we get

$\angle BCA = \angle DCA$
Therefore ,

$AC$ Bisects

$\angle DCB$ .
1746087_1831079_ans_acc55de47d794862b800dc7ec5b25356.png

Prove that
$\angle B = \angle C$

From the figure

$BD = CD$
In

$\Delta ABD$ and

$\Delta ACD$

$BD = CD$

$AD = AD$ is common

$\angle ADB=\angle ADC=90^o$

$\Delta ABD \equiv \Delta ACD$ (SAS Axiom)

$\angle B=\angle C$ (c.p.c.t)
Therefore , it is proved.
1746138_1831094_ans_6da748ee6ad94846a40961d670c45f1d.png

In the given figure , prove that :
$\Delta ADB \cong \Delta BCA$

and

It is given that

and

are vertically opposite angles

It is given that

is common
Here

Therefore , it is proved.

In the given figure , prove that :
$AD = BC$

$\Delta AOD \ \& \ \Delta BOC$

$OA = OB$ (given)

$OD = OC$ (given)

$\angle AOD = \angle BOC$ ( vertically opposite angles )

$\Delta AOD \cong \Delta BOC$ (by

$SAS$ condition of congruency )

$\implies AD = BC$ (corresponding parts of congruent triangles are equal)
Hence , it is proved.

In the given figure, prove that :
$\Delta AOD \cong \Delta BOC$

$\Delta AOD$ and

$\Delta BOC$ ,
It is given that:

$OA = OB$

$OD = OC$

$\angle AOD = \angle BOC$ {vertically opposite angles}

$\Delta AOD \cong \Delta BOC$ (SAS Axiom)
Therefore, it is proved.

In the given figure , prove that :
$\Delta XYZ \cong \Delta XPZ$

$\Delta XYZ$ and

$\Delta XPZ$
It is given that

$XY = XP$

$XZ = XZ$ is common

$\angle XYZ=\angle XPZ=90^o$

$\Delta XYZ \cong \Delta XPZ$ (SSA Axiom)
Therefore , it is proved .

In the given figure , prove that :
$YZ = PZ$

$\Delta XYZ$ and

$\Delta XPZ$
It is given that

$XY = XP$

$XZ = XZ$ is common

$\angle XYZ=\angle XPZ=90^o$

$\Delta XYZ \cong \Delta XPZ$ (SSA Axiom)

$YZ=PZ$ (c.p.c.t)
Therefore , it is proved .

The given figures shows a triangle ABC in which AD is perpendicular to side BC and BD = CD. Prove that :
(i) $\Delta ABD \cong \Delta ACD$
(ii) $AB = AC$
(iii) $\angle B = \angle C$

(i) In

$\Delta ABC$

$AD$ is perpendicular to

$BC$

$BD = CD$
In

$\Delta ABD$ and

$\Delta ACD$

$AD = AD$ is common

$\angle ADB = \angle ADC = 90^{\circ}$
It is given that

$BD = CD$

$\Delta ABD \cong \Delta CAD$ (SAS Axiom)
(ii)

$AB = AC$ (c.p.c.t)
(iii)

$\angle B = \angle C$ as

$\Delta ADB \cong \Delta ADC$
Therefore , it is proved.

In the given figure , $\angle 1 = \angle 2$ and $AB = AC$ .
Prove that:
(i) $\angle B = \angle C$
(ii) $BD = DC$
(iii) $AD$ is perpendicular to $BC$

$\Delta ADB$ and

$\Delta ADC$
It is given that

$AB = AC$
and

$\angle 1 = \angle 2$

$AD = AD$ is common
Hence ,

$\Delta ADB \cong \Delta ADC$ (SAS Axiom)

(i)

$\angle B = \angle C$ (c.p.c.t)

(ii)

$BD = DC$ (c.p.c.t)

(iii)

$\angle ADB = \angle ADC$ (c.p.c.t)
We know that

$\angle ADB + \angle ADC = 180^{\circ}$ is a linear pair
So we get

$\angle ADB = \angle ADC = 90^{\circ}$

$AD$ is perpendicular to

$BC$
Therefore , it is proved.

In the given figure , prove that :
$\angle YXZ = \angle PXZ$

$\Delta XYZ$ and

$\Delta XPZ$
It is given that

$XY = XP$

$XZ = XZ$ is common

$\angle XYZ=\angle XPZ=90^o$

$\Delta XYZ \cong \Delta XPZ$ (SSA Axiom)

$\angle YXZ = \angle PXZ$ (c.p.c.t)
Therefore it is proved.

In the given figure , prove that :
$\angle ADB = \angle ACB$

$\Delta AOD$ and

$\Delta BOC$
It is given that

$OA = OB$

and

$OD = OC$

$\angle AOD = \angle BOC$ are vertically opposite angles

$\Delta AOD \cong \Delta BOC$ (SAS Axiom)

$\angle ADB=\angle ACB$ (c.p.c.t)
Therefore , it is proved.

In the given figure , prove that :
$\Delta ABC \cong \Delta DCB$

$\Delta ABC$ and

$\Delta DCB$

$CB = CB$ is common
It is given that

$\angle ABC = \angle BCD = 90^{\circ}$

$AB = CD$

$\Delta ABC \cong \Delta DCB$ (SAS Axiom)
Therefore it is proved.

In the given figure $AB , = DB$ and $AC = DC$
If $\angle ABD = 58 ^{o}$
$\angle DCB = (2x - 4) ^{o}$
$\angle ACB = y + 15$ and
$\angle DCB = 63 ^{o}$ : find the values of $x$ and $y$

Given ; In the figure

$AB = DB , AC = DE$ ,

$\angle ABD = 58 ^{o}$

$\angle DBC = (2x - 4) ^{o}$ ,

$\angle ACB = (y + 15)$ and

$\angle DCB = 63 ^{o}$
We need find the values of

$x$ and

$y$
In

$\Delta ABC$ and

$\Delta DBC$

$AB = DB$ [given]

$AC= DC$ [given ]

$BC = BC$ [ common]
By side - side - side criterion of congruent , we have

$\Delta ABC \cong \Delta DBC$
The corresponding parts of the congruent triangles are congruent

$\angle ACB = \angle DCB [c . p . c. t ]$

$\Rightarrow y^{0} + 15^{0} = 63^{o}$

$\Rightarrow y^{0} = 63^{o} - 15^{o}$

$\Rightarrow y^{0} = 48 ^{o}$
and

$\angle ABC = \angle DBC [c.p.c.t]$

$\angle DBC = (2x - 4)^{o}$
We have

$\angle ABC + \angle DBC = \angle ABD$

$\Rightarrow 4x - 8 ^{o} - 58^{o}$

$\Rightarrow 4x - 8^{0}+ 8^{o}$

$\Rightarrow 4x - 66^{o}$

$\Rightarrow x \dfrac{66^{o}}{4}$

$\Rightarrow x = 16 . 5^{o}$
Thus the values of

$x$ and

$y$ are
;

$x = 16.5^{o} and y = 48^{o}$

In the following example, a pair of triangles is shown. Equal parts of triangle in each pair are marked with the same sign. Observe the figure and state the test by which triangles in each pair are congruent.

By.............test
$\Delta PQR \cong \Delta STU$

$\triangle PQR$ and

$\triangle STU$

$\angle P=\angle S$

$PR=ST$

$\angle R=\angle T$

$ASA$ Test

$\Delta PQR \cong \Delta STU$

1820354_1855904_ans_a06d8324f10d41c3ab62729c9e423e86.png

In the following example, a pair of triangles is shown. Equal parts of triangle in each pair are marked with the same sign. Observe the figure and state the test by which the triangle in each pair are congruent.

By............test
$\Delta LMN \cong \Delta PTR$

Hypotenuse Side Test

In the given figure, $\angle P =\angle R$ and $PQ =RQ.$ Prove that, $\Delta PQT \cong \Delta RQS$

$\Delta PQT$ and

$\Delta RQS$

$\angle P = \angle R$ (Given)

$PQ = RQ$ (Given)

$\angle Q = \angle Q$ (Common)

So, by

$ASA$ rule of congruence,

$\Delta PQR \cong \Delta RQS$

Observe the information shown in pair of triangle shown above. State the test by which the two triangles are congruent. Write the remaining congruent parts of the triangle.

Given is a figure having two triangles,

$\triangle PQT$ and

$\triangle SRT$

We can see from the figure that.

$PT=ST$

and

$QT=RT$

Also,

$\angle PTQ$ and

$\angle RTS$ are vertically opposite angles.

Hence,

$\angle PTQ = \angle RTS$

$\therefore \$ By

$SAS$ congruence criterion,

$\triangle PQT\cong \triangle SRT$

Now, since

$\triangle PQT$ and

$\triangle SRT$ are congruent, their corresponding sides and angles are equal.

$\therefore \ PQ=RS$

$\angle PQT=\angle SRT$

and

$\angle TPQ=\angle TSR$

Hence,

$\triangle PQT$ and

$\triangle SRT$ are congruent and their corresponding sides and angles are equal.

In the given alongside figure, $AD=AB=AC,BD$ is parallel to $CA$ and $\angle ACB=65^o$ .
Find the $\angle DAC$ .

We can see that

$\Delta ABC$ is an triangle with

$AB= AC$ .

$\Rightarrow \angle ACB=\angle ABC$
As

$\angle ACD=65^o$
hence,

$\angle ABC=65^o$
Sum of all the angles of a triangle is

$180^o$

$\angle ACB+\angle CAB+\angle ABC=180^o$

$65^o+65^o+\angle CAB=180^o$

$\angle CAB=180^-130^o$

$\angle CAB=50^o$
As

$BD$ is parallel to

$CA$
Therefore,

$\angle CAB=\angle DBA$ since they are alternate angles.

$\angle CAB=\angle DBA=50^o$
We see that

$\Delta ADB$ is an isosceles triangle with

$AD=AB$

$\Rightarrow \angle ADB=\angle DBA=50^o$
Sum of all the angles of a triangle is

$180^o$

$\angle ADB+\angle DAB+\angle DBA=180^o$

$50^o+\angle DAB+50^o=180^o$

$\angle DAB=180^o-100^o=80^o$

$\angle DAB=80^o$
The angle

$DAC$ is sum of angle

$DAB$ and

$CAB$

$\angle DAC=\angle CAB=\angle DAB$

$\angle DAC=50^o+80^o$

$\angle DAC=130^o$

As shown in the following figure, in $\Delta LMN$ and $\Delta PNM$ , $LM = PN, LN = PM.$ Write the test which assures the congruence of the two triangles. Write their remaining congruent part.

$\Delta LMN$ and

$\Delta PNM$

$LM = PN$ (Given)

$LN = PM$ (Given)

$MN = NM$ (Common)
By SSS test of congruency

$\Delta LMN \sim \Delta PNM$

$\angle L = \angle P$ (corresponding angles of congruent triangles)

$\angle LMN = \angle PNM$ (corresponding angles of congruent triangles)

$\angle MNL = \angle NMP$ (corresponding angles of congruent triangles)

Observe the information shown in the pair of triangles given below. State the test by which the two triangles are congruent. Write the remaining congruent parts of the triangles.

From the information shown in the figure,
In

$\Delta ABC$ and

$\Delta PQR$

$\angle ABC \cong \angle PQR$
seg

$BC \cong$ seg

$QR$

$\angle ACB \cong \angle PQR$

$\therefore \Delta ABC \cong \Delta PQR ....... by \ ASA$ test

$\therefore \ angle BAC \cong \angle QPR$ ....... corresponding angles of congruent triangles
seg

$AB \cong$ seg

$PQ$ , seg

$AC \cong$ seg

$PR$ }.....corresponding sides of congruent triangles

In the following example, a pair of triangles is shown. Equal parts of the triangle in each pair are marked with the same sign. Observe the figure and state the test by which the triangle in each pair are congruent.

By............test
$\Delta XYZ \cong \Delta LMN$

$\Delta XYZ$ and

$\Delta LMN$

$XY=LM$

$\angle Y=\angle M$

$YZ=MN$

$SAS$ Test

$\Delta XYZ \cong \Delta LMN$

In the figure, point $D$ and $E$ are on side $B C$ of $\Delta A B C,$ such that $B D=C E$ and $A D=A E$
show that $\triangle ABD \cong \triangle ACE$ .

$\ln \Delta \mathrm{ADE}, \mathrm{AD}=\mathrm{AE}$

$\therefore \angle \mathrm{ADE}=\angle \mathrm{AED} \quad$ (Angles opposite to equal sides)
Now,

$\angle \mathrm{ADE}+\angle \mathrm{ADB}=180^{\circ} \ldots$ (1)

$\angle \mathrm{AED}+\angle \mathrm{AEC}=180^{\circ} \ldots$ (2)
Subtracting (2) from

$(1),$ we get

$\angle \mathrm{ADB}-\angle \mathrm{AEC}=0 \ldots$ (3)

$\Rightarrow \angle A D B=\angle A E C$

$\ln \triangle \mathrm{ABD}$ and

$\triangle \mathrm{ACE}$

$\angle \mathrm{ADE}=\angle \mathrm{AEC} \quad[$ From (3)

$]$

$B D=C E \quad$ (Given)

$A D=A E \quad$ (Given)
By SAS test of congruency

$\triangle \mathrm{ABD} \cong \triangle \mathrm{ACE}$

Fill in the blank using the correct word given in brackets:
All squares are _____. (similar, congruent )

Similar,

because

all squares are of same shape but

size may not be same

Hence, these are similar but not congruent.

All circles are similarAll squares are similar.
1959745_1867752_ans_79d1d5ab1aff44acac8c05827f2e8dd1.png

Line 1 is the bisector of an angle $\angle A$ and B is any point on l. BP and BQ are perpendiculars from B to the arms of $\angle A$ . Show that :
(i) $\triangle APB \cong \triangle AQB$
(ii) BP = BQ or B is equidistant from the arms of $\angle A$ .

Data : Line

$I$ is the bisector of an angle

$\angle A$ and

$B$ is any point on

$I$ .

$BP$ and

$BQ$ are perpendiculars from

$B$ to the arms of

$\angle A$ .
To Prove:
(i)

$\Delta APB = \Delta AQB$
(ii)

$BP = BQ$ or

$B$ is equidistant from the arms of

$\angle A$ .
Proof: In

$\Delta APB$ and

$\Delta AQB$ ,

$\angle APB = \angle AQB = 90^0$

$\angle PAB = \angle QAB$ (

$\because I$ bisects

$\angle A$ ).

$AB$ is common.

$\therefore \ \ \Delta APB \cong \Delta AQB$ (

$ASA$ postulate)

$\therefore \ \ BP = BQ$ .

$B$ is equidistant from the arms of

$\angle A$ .

In the given figure, bisector of $\angle \mathrm{BAC}$ intersects side $\mathrm{BC}$ at point $\mathrm{D}$ . Prove that $\mathrm{AB}>\mathrm{BD}$

Given:

$\mathrm{AD}$ is the bisector of

$\angle \mathrm{A}$

$\therefore \angle D A B=\angle D A C$
Now,

$\angle \mathrm{BDA}$ is the exterior angle of the

$\Delta \mathrm{DAC}$

$\therefore \angle \mathrm{BDA}>\angle \mathrm{DAC}$

$\Rightarrow \angle B D A>\angle D A B \quad(\because \angle D A C=\angle D A B)$

$\Rightarrow \mathrm{AB}>\mathrm{BD} \quad$ (The sides opposite to the greater angle is greater than the side opposite to the smaller angle)
Hence proved.

In the given figure, point $S$ is any point on side $\mathrm{QR}$ of $\Delta \mathrm{PQR}$
Prove that $: \mathrm{PQ}+\mathrm{QR}+\mathrm{RP}>2 \mathrm{PS}$

$\Delta PQS$

$\mathrm{PQ}+\mathrm{QS}>\mathrm{PS}$

$\ldots$ (1) (Sum of two sides of a traingle is greater than the third side)

$\ln \Delta \mathrm{PRS}$

$\mathrm{RP}+\mathrm{RS}>\mathrm{PS} \ldots$ (2) (Sum of two sides of a traingle is greater than the third side)
Adding (1) and

$(2),$ we get

$\mathrm{PQ}+\mathrm{QS}+\mathrm{RP}+\mathrm{RS}>\mathrm{PS}+\mathrm{PS}$

$\Rightarrow \mathrm{PQ}+\mathrm{QR}+\mathrm{RP}>2 \mathrm{PS}$ Proved

In the figure, points D and E are on side BC of $\Delta$ ABC, such that BD = CE and AD = AE.

Show that $\Delta \mathrm{ABD} \cong \Delta \mathrm{ACE}$

$\Delta \mathrm{ADE, AD = AE}$

$\therefore \angle \mathrm{ADE} = \angle \mathrm{AED}$ (Angles opposite to equal sides)
Now,

$\angle \mathrm{ADE} + \angle \mathrm{ADB} = 180^{\circ}$ ...........(1)

$\angle \mathrm{AED} + \angle \mathrm{AEC} = 180^{\circ}$ ...........(2)
Subtracting (2) from (1), we get

$\angle \mathrm{ADE} + \angle \mathrm{ADB} = 0$

$\Rightarrow \angle \mathrm{ADB} = \angle \mathrm{AEC}$ ......(3)
In

$\Delta \mathrm{ABD}$ and

$\Delta \mathrm{ACE}$

$\Rightarrow \angle \mathrm{ADE} = \angle \mathrm{AEC}$ ......[From (3)]

$\mathrm{BD = CE}$ (Given)

$\mathrm{AD = AE}$ (Given)
By SAS test of congruency

$\Delta \mathrm{ABD} \cong \Delta \mathrm{ACE}$

In $\Delta \mathrm{PQR},$ If $\mathrm{PQ}>\mathrm{PR}$ and bisectors of $\angle \mathrm{Q}$ and $\angle \mathrm{R}$ intersect at $\mathrm{S}$ . Show that $\mathrm{SQ}>\mathrm{SR}$ .

$\ln \Delta \mathrm{PQR}, \mathrm{PQ} >\mathrm{PR}$
Now, the angle opposite to the greater side is greater than angle opposite to the smaller side.

$\therefore \angle R>\angle Q$
Dividing both sides by

$2,$ we get

$\dfrac{1}{2} \angle \mathrm{R}>\dfrac{1}{2} \angle \mathrm{Q}$

$\Rightarrow \angle S R Q>\angle S Q R$

$\Rightarrow \mathrm{SQ}>\mathrm{SR} \quad$ (The sides opposite to the greater angle is greater than side opposite to the smaller angle)
Hence proved.

Fill in the blank using the correct word given in brackets:
All circles are ______. (Congruent, similar)

Similar

because all circles are of same shape (round) but

size may not be same

Hence, these are similar but not congruent.

All circles are similar
1959742_1867751_ans_7691be33b32a4174aebdfc2c2e884851.png

l and m are two parallel lines intersected by another pair of parallel lines p and q. Show that $\triangle ABC \cong \triangle CDA$ .

$I$ and

$m$ are two parallel lines intersected by another pair of parallel lines

$p$ and

$q$ .
To Prove:

$\Delta ABC \cong \Delta CDA$
Proof: In

$\Delta ABC$ and

$\Delta CDA$ ,

$\angle ACB = \angle DAC \because$ Alternate angles.

$\angle BAC = \angle ACD$

$AC$ is common.

$A.S.A.$ postulate.

$\Delta ABC \cong \Delta CDA$

ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal. Show that
(i) $\triangle ABE = \triangle ACF$
(ii) $AB = AC$ i.e., $\triangle ABC$ is an isosceles triangle.

Data : In

$\Delta ABC$ , altitudes

$BE$ and

$CF$ to sides

$AC$ and

$AB$ are equal and

$BE = CF$ .
To Prove:
(i)

$\Delta ABE \cong \Delta ACF$
(ii)

$AB = AC$ i.e.,

$\Delta ABC$ is an isosceles triangle.
Proof: In

$\Delta ABE$ and

$\Delta ACF$ ,

$\angle AEB = \angle AFC = 90^0$ (Data)
Altitude

$BE$ = Altitude

$CF$ (Data)

$\angle A$ is Common.

$\therefore \ \ \Delta ABE \cong \Delta ACF$ (MS Postulate)

$\therefore \ AB = AC$

$\Delta ABC$ is an isosceles triangle.

In fig $\angle B < \angle A$ and $\angle C < \angle D$ . Show that AD < BC.

Data :

$\angle B < \angle A$ and

$\angle C < \angle D$ .
To Prove:

$AD < BC$
Proof:

$\angle B < \angle A$

$\therefore \ \ OA < OB.....(i)$
Similarly,

$\angle C < \angle D$ ,

$\therefore \ \ OD < OC....(ii)$
Adding

$(i)$ and

$(ii)$ , we have

$OA + OD < OB + OC$

$\therefore \ \ AD < BC$ .

In sides AB and AC of $\triangle ABC$ are extended to points P and Q respectively. Also, $\angle PBC < \angle QCB$ . Show that AC > AB.

Data :

$AB$ and

$AC$ are the sides of

$\Delta ABC$ and

$AB$ and

$AC$ are produced to

$P$ and

$Q$ respectively.

$\angle PBC < \angle QCB$ .
To Prove:

$AC > AB$
Proof:

$\angle PBC < \angle QCB$
Now,

$\angle PBC + \angle ABC = 180^0$

$angle ABC = 180 - \angle PBC.....(i)$
Similarly,

$\angle QCB + \angle ACB = 180^0$

$\angle ACB = 180 - \angle QCB....(ii)$
But,

$\angle PBC < \angle QCB$ (Data)

$\therefore \ \angle ABC > \angle ACB$ Comparing (i) and (ii),

$AC > AB$ (

$\because$ Angle opposite to larger side is larger)

Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of $\triangle PQR$ . Show that :
(i) $\triangle ABM \cong \triangle PQN$
(ii) $\triangle ABC \cong \triangle PQR$

Data : Two sides

$AB$ and

$BC$ and median

$AM$ of one triangle

$ABC$ are respectively equal to sides

$PQ$ and

$QR$ and median

$PN$ of

$\Delta PQR$ .
To Prove:
(i)

$\Delta ABM \cong \Delta PQN$
(ii)

$\Delta ABC \cong \Delta PQR$ .
Proof: (i) In

$\Delta ABC$ ,

$AM$ is the median drawn to

$BC$ .

$\therefore \ \ BM = \frac{1}{2}BC$
Similarly, in

$\Delta PQR$ ,

$QN = \frac{1}{2} QR$
But,

$BC = QR$

$\frac{1}{2} BC = \frac{1}{2} QR$

$\therefore \ \ BM = QN$
In

$\Delta ABM$ and

$\Delta PQN$ ,

$AB = PQ$ (data)

$BM = QN$ (data)

$AM = PN$ (proved)

$\therefore \ \ \Delta ABM \cong \Delta PQN$ (

$SSS$ postulate)
(ii) In

$\Delta ABC$ and

$\Delta PQR$ ,

$AB = PQ$ (data)

$\angle ABC = \angle PQR$ (proved)

$BC = QR$ (data)

$\Delta ABC \cong \Delta PQR$ (

$SSS$ postulate)

ABC and DBC are two isosceles triangles on the same base BC. Show that $\angle ABD = \angle ACD$ .

Data:

$ABC$ and

$DBC$ are two isosceles triangles on the same base

$BC$ .
To Prove:

$\angle ABD = \angle ACD$
Proof: In

$\Delta ABC, AB = AC$ ,

$\therefore$ Opposite angle

$\angle ABC = \angle ACB$
(i) Similarly in

$\Delta BDC, BD = DC$ .
Opposite angle

$\angle DBC = \angle DCB$
(ii) From (i) and (ii),

$\angle ABC = \angle ACB$
Adding

$\angle DBC$ and

$\angle DCB$ on both sides,

$\angle ABC + \angle DBC = \angle ACB+\angle DCB$

$\angle ACB + \angle DCB = \angle ACD$
Equals are added to equal angles.

$\therefore \ \ \angle ABD = \angle ACD$ .

AB is a line segment and P is its midpoint. D and E are points on the same side of AB such that $\angle BAD = \angle ABE$ and $\angle EPA = \angle DPB$ . Show that
(i) $\triangle DAP \cong \triangle EBP$
(ii) AD = BE

Data :

$AB$ is a line segment and

$P$ is its mid-point.

$D$ and

$E$ are points on the same side of

$AB$ such that

$\angle BAD = \angle ABE$ and

$\angle EPA = \angle DPB$ .
To Prove :
(i)

$\Delta DAP \cong \Delta EBP$
(ii)

$AD = BE$
Proof : (i) In

$\Delta DAP$ and

$\Delta EBP$ ,

$AP = BP$ (

$\because \ \ P$ is the mid-point of

$AB$ )

$\angle BAD = \angle ABE$ (Data)

$\angle APD = \angle BPE$

$\because$ (

$\angle EPA = \angle DPB$ Adding

$\angle EPD$ to both sides)

$\angle EPA + \angle EPD = \angle DPB + \angle EPD$

$\therefore \ \ \angle APD = \angle BPE$ .
Now, Angle, Side, Angle postulate.

$\therefore \ \ \Delta DAP \cong \Delta EBP$ .
(ii) As it is

$\Delta DAP \cong \Delta EBP$ ,
Three sides and three angles are equal to each other.

$\therefore \ \ AD = BE$ .

In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B. Show that :
(i) $\triangle AMC \cong \triangle BMD$
(ii) $\angle DBC$ is a right angle.
(iii) $\triangle DBC \cong \triangle ACB$
(iv) $CM = \frac{1}{2} AB$ .

Data given:

$\angle{ACB}={90}^{0}$

$M$ is the midpoint of

$AB$

$AM=BM$ .....

$\left(1\right)$

Also,

$DM=CM$ ......

$\left(2\right)$

$\left(A\right)$ Lines

$CD$ and

$AB$ intersect at

$M$ .

So,

$\angle{AMC}=\angle{BMD}$ (vertically opposite angles) ......... .

$\left(3\right)$

$\triangle AMC$ and

$\triangle BMD$

$\triangle AMC\cong \triangle BMD$

$AM=BM$ (from

$\left(1\right)$ )

$\angle{AMC}=\angle{BMD}$ (from

$\left(3\right)$ )

$CM=DM$

$\therefore \triangle AMC\cong \triangle BMD$ (by SAS congruence rule)

$\therefore \angle{ACM}=\angle{BDM}$ (by C.P.C.T)

But

$\angle{ACM}$ and

$\angle{BDM}$ are alternate interior angles for lines

$AC$ and

$BD$ .

If a transversal intersects two lines such that pair of alternate interior angles is equal, then the lines are parallel.

$BD\parallel AC$

Since

$BD\parallel AC$ and considering

$BC$ as transversal.

$\Rightarrow \angle{DBC}+\angle{ACB}={180}^{0}$ since interior angles on the same side of transversal are supplementary.

$\Rightarrow \angle{DBC}+{90}^{0}={180}^{0}$

Hence

$\angle{DBC}$ is a right angle

From

$\left(1\right) \triangle AMC\cong\triangle BMD$

$\Rightarrow AC=BD$ by C.P.C.T .....

$\left(4\right)$

$\triangle DBC$ and

$\triangle ACB$ ,

$DB=AC$ from

$\left(4\right)$

$\angle{DBC}=\angle{ACB}$ both

${90}^{0}$

$BC=CB$ (common)

$\therefore \triangle DBC\cong \triangle ACB$ (SAS congruence rule)

$\therefore DC=AB$ by C.P.C.T

$\Rightarrow \dfrac{1}{2}DC =\dfrac{1}{2}AB$ as

$DM=CM$

$\therefore DM=CM=\dfrac{1}{2}DC$

$\therefore CM=\dfrac{1}{2}AB$

1826954_1868970_ans_044971a0c20847d3b907c9560e83c060.png

In AC = AE, AB = AD and $\angle BAD = \angle EAC$ . Show that BC = DE

Data:

$AC = AE, AB = AD$ and

$\angle BAD = \angle EAC$
To Prove :

$BC = DE$
Proof : In

$\Delta ABC$ and

$\Delta EAD$ ,

$AB = AD$ (Data)

$AC = AE$ (Data)

$\Delta BAC = \Delta EAD$
(

$\because \ \ \angle BAD + \angle DAC = AEAC + \angle DAC$ and

$\angle DAC = \angle DAC.)$
Side - Angle - Side (

$SAS$ ) Postulate

$\therefore \ \ \Delta ABC = \Delta EAD$

$\therefore \ \ BC = DE$ .

$ABC$ is an isosceles triangle with $AB = AC$ . Show that $\angle B = \angle C$ .

Given,

$ABC$ is an isosceles triangle with

$AB = AC$ .
To Prove :

$\angle B = \angle C$
Construction: Draw

$AP \perp BC$ .

Proof: In

$\Delta ABC, AP \perp BC$ and

$AB = BC$ .

$\therefore$ In

$\Delta ABP$ and

$\Delta ACP$

$\angle APB = \angle APC = 90^0 (\because AP \perp BC)$
Hypotenuse

$AB$ = Hypotenuse

$AC$

$AP$ is common.
As per

$RHS$ postulate,

$\Delta ABP \cong \Delta ACP$

$\therefore \ \angle ABP = \angle ACP$ [by CPCT]

$\therefore \ \ \angle ABC = \angle ACB$

$\therefore \ \angle B = \angle C$ .

$(Hence\ proved)$

BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.

$\textbf{Step 1: Considering triangle BCF and CBE.}$

$\text{In }\Delta BCF \text{ and }\Delta CBE.$

$\angle BFC=\angle CEB (\text{Each }90^\circ)$

$\text{Hypotenuse BC}=\text{Hypotenuse BC} (\text{Common Side})$

$\text{Side FC = Side EB (Given )}$

$\therefore \text{By R.H.S criterion of congruence, we have }$

$\Delta \text{BCF} \cong \Delta \text{CBE}$

$\therefore \angle \text{ FBC}=\angle\text{ ECB (Corresponding Part of Congruent Triangles)}$

$\textbf{Step 2: Considering triangle ABC.}$

$\text{In }\Delta \text{ABC},$

$\angle\text{ ABC}=\angle\text{ ACB}$

$[\because \angle\text{ FBC}=\angle\text{ ECB}]$

$\implies\text{AB = AC}$

$\implies \Delta \text{ABC} \text{ is an isosceles triangle}$

$\textbf{Hence, it is proved that the given triangle ABC is an isosceles traingle.}$

In ( $\triangle ABC , AB = BC$ and $| B = 64^{\circ}$ find | $\angle C$

$AB = BC$ [data ]

$\therefore \angle C = \angle A$ [ Theorem | ]

$\angle A + \angle B + \angle C = 180^{\circ}$
( Sum of the angle of a triangle is

$180^{\circ}$ )

$\angle C + 64 + \angle C = 180^{\circ} [ \angle A = \angle C ]$

$64 + 2\angle C = 180^{\circ}$

$2 \angle C = 116$

$\angle C = \dfrac{116}{2} = 58^{\circ}$

In a triangle $ABC , AB = AC$ . Points G on AB and D on AC are such that $AE = AD$ prove that triangles BCD and CBE are congruent

In the figure

$AB = AC$ (data )

$\therefore \angle ABC = \angle ACB$
[ Base angles of an isosceles triangles ]

$AB = AC$ (data )

$AE = AD$ (data )

$AB - AE = AC - AD$ (Axiom 3)

$BE = DC$
In

$\bigtriangleup BCD$ and

$\bigtriangleup CBE$

$\angle BCD = \angle EBC$

$BE = DC$ (proved )

$BC = BC$ (Common side )

$\therefore \bigtriangleup BCD = \bigtriangleup CBE$ [ S AS postulate ]

Justify the following statement with reason Difference of any two side is less than the third side

$\bigtriangleup ABC$
To prove (i)

$AC - AB < BC (ii) BC - AC < AB (iii) BC - AB < AC$
Construction Let

$AC < AB .$ On AC
Proof :

$AB = AD$

$\therefore \angle 1 = \angle 2$ [ Base angles of isosceles triangle]
Extr

$\angle 2 = \angle 4 + \angle 3$

$\angle 2 > \angle 4$ [ Exterior angle > each interior opp angle ]

$\therefore$ Extr

$\angle 3 = \angle 1 + \angle A$
1986142_1871241_ans_dd226eed03384bdaa5a56a4473d0f5a0.png

In right triangle the make the statement true
The sum of any two sides of a triangle is ____ than the third side

1972290_1871205_ans_52c603dd9b6d426eae60beba8161213e.png

In $\bigtriangleup ABC , AC = AB$ and the altitude AD bisects BC prove that $\bigtriangleup ADC = \bigtriangleup ADB$

$\bigtriangleup ABD$ and

$\bigtriangleup ACD$

$AB = AC$ [ data]

$BD = DC$ [data]

$AD = AD$ [Common side]

$\therefore \bigtriangleup ABD = \bigtriangleup ACD$ [ SSS postulate]

In Fg. PR > PQ and PS bisects $\angle QPR$ . Prove that $\angle PSR > \angle PSQ$ .

Data :

$PR > PQ$ and

$PS$ bisects

$\angle QPR$ .
To Prove:

$\angle PSR > \angle PSQ$
Proof : In

$\Delta PQR, PR > PQ$ .

$\therefore \ \ \angle PQR > \angle PRQ ......(i)$

$PS$ bisects

$\angle QPR$ .

$\therefore \ \ \angle QPS = \angle RPS.....(ii)$
By adding (i) and (ii),

$\angle PQR + \angle QPS > \angle PRQ + \angle RPS....(iii)$
But, in

$\Delta PQS, \angle PSR$ is exterior angle.

$\therefore \ \ \angle PSR = \angle PQR + \angle QPS.....(iv)$
Similarly, in

$\Delta PRS, \angle PSQ$ is exterior angle.

$\therefore \ \angle PSQ = \angle PRS + \angle RPS...(v)$
In

$(iii)$ , subtracting

$(iv)$ and

$(v)$ ,

$\angle PSR > \angle PSQ$ .

Let ABC be a triangle and P be an interior point , prove that $AB + BC + CA < 2 ( PA + PB + PC )$

$\bigtriangleup PBA , AB < PA + PB ....(i)$
In

$\bigtriangleup PBC , BC < PB + PC ....(ii)$
In

$\bigtriangleup PCA , AC < PC + PA ....(iii)$

$AB + BC + AC < PA + PB + PB + PC + PA > 2PB$

$2PC + AB BC + AC< 2 ( PA + PB + PC )$

In the adjoining figure If $BC = EF$ then, $\angle C =$ _ and $\angle A =$ ___

$\angle C = \angle F$ and

$\angle A = \angle D$

Let AABC be a triangle such that $\angle B = 70^{\circ}$ and $\angle C = 40^{\circ}$ . Suppose D is a point on BC such that $AB = AD$ Prove that $Ab > CD$

$\bigtriangleup ABD, AB = AD$

$\therefore \angle ABD = \angle APB = 70^{\circ}$ [ Theroem |]

$\angle APB + \angle ADC = 180^{\circ}$ [ Linear pair]

$70 + \angle ADC = 180^{\circ}$

$\angle ADC = 180 - 70^{\circ}$

$\angle ADC = 180 - 70$

$\angle ADC = 110^{\circ}$
In

$\bigtriangleup ADC$

$\angle DAC = 180 - (110 + 40)$

$= 180 - 150$

$\angle DAC = 30^{\circ}$
In

$\bigtriangleup ADC$

$\angle ACD > \angle DAC$

$AD > CD$

$\therefore AB > CD$

In a $\bigtriangleup ABC , AB = AC$ and $| A = 50^{\circ}$ find $\angle B$ and $\angle C$

$\angle A + \angle B +\angle C = 180^{\circ}$
(Sum of the angle of a triangle is

$180^{\circ}$ )

$50 + \angle B + \angle B = 180^{\circ}$

$\because \angle B = \angle C$ Base angles of an isosceles triangle

$50 + 2 \angle B = 180^{\circ}$

$2 \angle B =- 180 - 50$

$2 \angle B = 130^{\circ}$

$\angle B = \dfrac{130}{2}$

$\angle B = 65^{\circ}$

$\angle B = \angle C = 65^{\circ}$

Two triangles $ABC$ and $DBC$ have common base $BC$ . Suppose $AB= DC$ and $\angle ABC = \angle BCD$ , then prove that $AC = BD$ .

$\bigtriangleup ABC$ and

$\bigtriangleup DBC$

$AB = DC$ (Given)

$\angle ABC = \angle BCD$ (Given)

$BC = BC$ (common side)

$\therefore \bigtriangleup ABC \cong \bigtriangleup DBC$ (By

$SAS$ postulate)

$\implies AC = BD$ ( Corresponding sides of congruent triangles)

1956520_1871256_ans_7971358c26b74cbb839ffd6df1a04917.png

In triangles, $ABC$ and $PQR, \angle A = \angle Q$ and $\angle B = \angle R$ then which side of $\Delta PQR$ is equal to side $BC$ of $\Delta ABC$ , so that both triangles congruent? Give reasons for your answer.

$\Delta ABC$ and

$\Delta PQR$

$\angle A = \angle Q$

$\angle B = \angle R$

$\angle C = \angle P$
(when two angles of

$\Delta ABC$ and

$\Delta PQR$ are equal then third angle of

$\Delta ABC$ automatically equal to third angle of other

$\Delta PQR$ )

$PR = BC$

$\Delta ABC \cong \Delta PQR$ (By ASA)

If two sides and one angle of a triangle are equal to two sides and the angle of other then both the triangles must congruent to each other. Is this statement true?

No, for any two triangles to be congruent it is must that any two sides and included angle of one is correspondingly equal to the other.

Let ABC be a triangle . Drawn a triangle BDC externally on BC such that AB = BD and AC = CD Prove that $\bigtriangleup ABC \cong \bigtriangleup DBC$

$\bigtriangleup ABC$ and

$\bigtriangleup DBC$

$AB = BD$ [data]

$AC = CD$ [ data]

$BC = BC$ [ common side ]

$\therefore \bigtriangleup ABC = \triangle DBC$ [ SSS postulate]
1956517_1871287_ans_64d823cf51e44033a51fd52e40ba2da1.png

In figure, $ADBC$ is a quadrilateral in which $\angle ABC = \angle ABD$ and $BC = BD$ then show that $\Delta ABC \cong \Delta ABD$ .

$\Delta ABC$ and

$\Delta ABD$

$\angle ABC = \angle ABD$ (given)

$BC = BD$ (given)

$AB = AB$ (common side)
similarly,

$\Delta ABC \cong \Delta ABD$
(By SAS congruency property)
Hence proved.

In $\Delta ABC$ and $\Delta PQR$ , $\angle A = \angle Q$ and $\angle B = \angle R$ . Then which side of $\Delta PQR$ is equal to side $AB$ of $\Delta ABC$ . So that both triangles become congruent. Give reason for your answer.

$\Delta ABC$ and

$\Delta PQR$

$\angle A = \angle Q, \angle B = \angle R$ then side

$AB$ should equal to

$QR$
i.e.,

$AB = QR$

$\therefore \Delta AOC \cong \Delta PQR$ (by ASA)

If two angles and one side of a triangle are equal to two angles and one side of other triangle then both As must congruent to each other. Is this statement true?

No, side must be included between the angles.

In the adjoining figure If $AC = CE$ and $\bigtriangleup ABC \cong \bigtriangleup DEC$ then $\angle D =$ _ and $\angle A =$ ___

1972277_1871354_ans_d6a1f7908ac542769aefac5781da0492.jpg

In figure, $Q$ is a point on side $SR$ of $\Delta PSR$ such that $PQ = PR$ . Prove that $PS > PQ$ .

$\Delta PQR$

$PQ = PR$

$\angle PQR = \angle PRQ$ …(i) (angles opposite to equal sides of a triangle are equal)
In

$\Delta PSQ$
Ext.

$\angle PQR > \angle PSQ$ …(ii)
From (i) and (ii), we get

$\angle PRQ > \angle PSQ$

$PS > PR$ (side opposite to greater angle is longer)

$PS > PQ (\because PQ = PR)$

$\Delta ABC$ and $\Delta DBC$ are two isosceles $I$ triangles on the same base $BC$ and vertices $A$ and $D$ are on the same side of $BC$ (see figure). If $AD$ is extended to intersect $BC$ at $P$ , show that
(i) $\Delta ABD \cong \Delta ACD$
(ii) $\Delta ABP \cong \Delta ACP$
(iii) $AP$ bisects $\angle A$ as well as $\angle D$
(iv) $AP$ is the perpendicular bisector of $BC$ .

(i) In

$\Delta ’s$

$ABD$ and

$ACD$ , we have

$AB = AC$ (given)

$BD = DC$ (given)
and

$AD = AD$ (common)

$\therefore \Delta ABD \cong \Delta ACD$
(by SSS congruency rule)
(ii) In

$\Delta ’s$

$ABP$ and

$ACP$ , we have

$AB = AC$ (given)

$\angle BAP = \angle CAP$
and

$AP = AP$ (common)

$[\because \Delta ABD \cong \Delta ACD, \angle BAD = \angle CAD, \angle BAP = \angle CAP]$

$\therefore \Delta ABP \cong \Delta ACP$
(by SAS congruency rule)
(iii) We have already proved in (i) that

$\Delta ABD \cong \Delta CAD$

$\angle BAP = \angle CAP$

$AP$ bisects

$\angle A$ i.e.

$AP$ is the bisector of

$\angle A$ .
In

$\Delta ’s$

$BDP$ and

$CDP$ , we have

$BD = CD$ (given)

$BP = CP [\because \Delta ABP = \Delta ACP]$
and

$DP = DP$ (common)

$\therefore \Delta BDP \cong \Delta CDP$
(by SSS congruency rule)

$\angle BDP = \angle CDP$

$DP$ is the bisector of

$\angle D$ .
Hence,

$AP$ is the bisector of

$\angle A$ as well as

$\angle D$ .
(iv) In (iii), we have proved that

$\Delta BDP \cong \Delta CDP$

$BP = CP$ and

$\angle BPD = \angle CPD = 90^{0}$ .

$\therefore \angle BPD$ and

$\angle CPD$ form a linear pair

$DP$ is the perpendicular bisector of

$BC$
Hence,

$AP$ is the perpendicular bisector of

$BC$

$AD$ is the median of any $\Delta ABC$ . Is it true to say that $AB + BC + CA > 2AD$ . Give reason for your answer.

Yes, if

$AD$ is median of

$\Delta ABC$

$AB + BD > AD$ …(i) (sum of any two sides of a

$A$ is greater than third side)
and

$AC + DC > AD$ …(iii) (reason as above)
Similarly,

$AB + (BD + DC) + AC > 2 AD$

$AB + BC + CA > 2AD$
1879518_1878696_ans_3c0cbbef5f4543f79eb2fa4c70dc611b.png

$BE$ and $CF$ are two equal altitudes of $\Delta ABC$ . By using RHS congruency rule, prove that $\Delta ABC$ is an isosceles triangle.

$\Delta BFC$ and

$\Delta CEB$

$\angle BFC = \angle CEB = 90^{0}$ (given)
hyp.

$BC$ = hyp.

$BC$ (common)
and altitude

$CF$ = altitude

$BE$

$\Delta BFC \cong \Delta CEB$
(by RHS congruency rule)

$\angle B = \angle C$

$\Delta ABC$ is an isosceles triangle.
Hence proved.
1869975_1878529_ans_04292e1b060540f59534dcadd7781f3a.png

In $\Delta ABC$ , $\angle A > \angle B$ and $\angle B > \angle C$ , then write the smallest side.

According to given relation

$\angle A > \angle C$ .
Side

$AB$ will be least because

$\angle C$ is least.

In an $\Delta ABC$ with $AB = AC$ , $D$ is any point on the side $AC$ . Show that $CD < BD$ .

$\Delta ABC$

$AB = AC$ (given)

$\angle B = \angle C$

$\angle ADB = \angle DBC + \angle C$ (exterior angle is equal to sum of opposite interior angles)

$\angle ADB > \angle C$

$AB > BD$
But

$AB = AC$

$AD + DC > BD$

$BD > CD$
Hence proved.
1869995_1878733_ans_dc9bd5ad201645058903d9a9e8c31382.png

In $\Delta PQR$ , $S$ is any point on side $QR$ . Show that $PQ + QR + RP > 2PS$ .

$\Delta PQR$ ,

$PQ + QS > PS$ …(i) (

$\because$ the sum of any two sides of a triangle is greater than the third side)
In

$\Delta PRS$ ,

$PR + RS > PS$ (

$\because$ The sum of any two sides of a triangle is greater than the third side)
On adding (i) and (ii), we get

$(PQ + QS) + (PR + RS) > 2PS$

$PQ + (QS + RS) + PR > 2PS$

$PQ + QR + PR > 2PS$

$PQ + QR + RP > 2PS$
Hence proved.
1869994_1878729_ans_d3dfd75f771a420bb76ec317449128d3.png

In figure, $AD = BD$ and $\angle C = \angle E$ . Prove that $BC = AE$ .

$\Delta AED$ and

$\Delta BCD$

$AD = BD$ (given)

$\angle ADE = \angle BDC$ (vertically opposite angles)

$\angle E = \angle C$

$\therefore \Delta AED \cong \Delta BCD$ (by AAS congruency property)
Similarly,

$AE = BC$ (by c.p.c.t)

In figure, $O$ is an interior point of $\Delta ABC$ . Show that: $AB + BC + CA < 2(OA + OB +OC)$ .

In triangle

$ABC$ ,

$O$ is a point interior of

$\Delta ABC$ .
As we know that “The sum of any two sides of a triangle is greater than the third side”.

$OA + OB > AB$ …(i)

$OA + OC > AC$ …(ii)
and

$OB + OC > BC$ …(iii)
Now, adding (i), (ii) and (iii), we get

$2(OA + OB + OC) > AB + BC + CA$
or

$AB + BC + CA < 2(OA + OB + OC)$
Hence proved.

In any triangle ABC if $AB > AC$ and $D$ is any point on BC then prove that $AB > AD$ .

$\Delta ABC$

$AB > AC$

$\angle ACB > \angle ABC$ …(i) (

$\because$ angle opposite to larger side is greater)
Now in

$\Delta ACD, CD$ is produced to

$B$ forming an ext.

$\angle ADB$ .

$\angle ADB > \angle ACD$ (exterior angle of a

$A$ is greater than each interior opposite angle)

$\angle ADB > \angle ACB$ …(ii)

$\angle ACD = \angle ACB$
From (i) and (ii)

$\angle ADB > \angle ABC$

$\angle ADB > \angle ABD [\because \angle ABC = \angle ABD]$

$AB > AD$ [

$\because$ side opposite to greater angle is larger]
Hence proved.
1869997_1878736_ans_e0b8e46b1a53446495d57842db9702dc.png

In the given figure, $O$ is the middle point of both $AB$ and $CD$ . Prove that $AC = BD$ and $AC || BD$ .

$\Delta ’s$

$AOC$ and

$BOD$

$OA = OB$

$OC = OD$

$\angle AOC = \angle BOD$ (vertically opposite angles)

$\Delta AOC = \Delta BOD$ (by SAS congruence property)
Hence

$AC = BD$ (by c.p.c.t)
Also

$\angle OAC = \angle OBD$ (alt. angle)

$AC || BD$
Hence proved.

In figure, $CB = AD$ and $AB = CD$ . Can we say $\angle ABC$ and $\angle ADC$ are equal? Why?

$\Delta ABC$ and

$\Delta ADC$

$AB = DC$

$AD = CB$
and

$AC = AC$

$\Delta ABC = \Delta ADC$ (by SSS congruency property)
So,

$\angle ABC = \angle ADC$ (by c.p.c.t)
Hence proved.

Prove that the difference of any two sides of a triangle is less than the third side.

Given: A

$\Delta ABC$
To prove:
(i)

$AC – AB < BC$
(ii)

$BC – AC < AB$
(iii)

$BC – AB < AC$
Construction: Take a point

$D$ on

$AC$ such that

$AB = AD$ . Join

$BD$ .
Proof: In

$\Delta ABD, AB = AD, \angle 1 = \angle 2$ …(i)
In

$\Delta ABD$ , side

$AD$ is produced to

$C \angle 3 > \angle 1$ …(ii)
(ext. angle is greater than one of the opposite interior angle)
Also in

$\Delta BCD$

$\angle 2 > \angle 4$ …(iii) (reason as above)
From (i) and (ii), we get

$\angle 3 > \angle 2$ …(iv)
From (iii) and (iv), we get

$\angle 3 > \angle 4$

$BC > CD$ (side opposite to greater angle is larger)

$CD < BC$

$AC – AD < BC$

$AC – AB < BC$

$[\because AD = AB]$
Similarly we can prove

$BC – AC < AB$ and

$BC – AB < AC$
Hence proved.
1870003_1878756_ans_03019c42472845a48ac750fc50b4cfe5.png

In figure, $\angle B > \angle A$ and $\angle D > \angle E$ then show that $AE > BD$ .

$\angle B > \angle A$

$AC > BC$ …(i)
Also

$\angle D > \angle E$

$CE > CD$ …(ii) (greater angle has larger side opposite to it)
Adding (i) and (ii),
we get

$AC + CE > BC + CD$

$AE > BD$
Hence proved.

In figure, $BA \perp AC, DE \perp EF$ , such that $BA = DE$ and $BF = CD$ , prove that $AC = EF$ .

To prove:

$AC = EF$ , we have to prove

$\Delta ABC = \Delta DEF$
Here

$BA = DE$ (given) …(i)

$BA \perp AC$ and

$DE \perp FE$

$\angle BAC = \angle DEF = 90^{0}$ (each) …(ii)
Also

$BF = CD$ (given) …(iii)
Adding

$FC$ in equation (iii), we get

$BF + FC = CD + FC$

$BC = FD$ …(iv)
From (i), (ii) and (iv), we get

$\Delta ABC = \Delta DEF$ (by RHS congruency property)

$AC = EF$ (by c.p.c.t)
Hence proved.

In figure, side $AB$ and $AC$ are produced to point $D$ and $E$ respectively. Bisectors $BO$ and $CO$ of $\angle DBC$ and $\angle ECB$ respectively meet at $O$ . If $AB > AC$ . Prove that $OC > OB$ .

Given:

$ABC$ is a triangle in which

$AB$ and

$AC$ is produced up to

$D$ and

$E$ respectively and bisectors of

$\angle DBC$ and

$\angle ECB$ meet at

$O$ . Also

$AB > AC$ .
To prove:

$OC > OB$
Proof: In

$\Delta ABC$

$AB > AC$

$\angle ACB > \angle ABC$ …(i)
Also

$BO$ and

$CO$ are the bisectors of

$\angle DBC$ and

$\angle BCE$ respectively.

$\angle OBD = \angle OBC$
and

$\angle OCB = \angle OCE$

$\angle ACB = 180^{0} – \angle BCE$

$\angle ACB = 180^{0} – 2\angle OCB$ …(ii)
Similarly

$\angle ABC = 180^{0} – 2 \angle OBC$ …(iii)
From (i), (ii) and (iii), we have

$180^{0} – 2\angle OCB > 180^{0} – 2\angle OBC$

$– 2\angle OCB > – 2\angle OBC$

$\angle OBC > \angle OCB$

$OC > OB$
Hence proved.

In the given figure, $PQRS$ is a quadrilateral. $PQ$ is its longest side and $RS$ is its shortest side. Prove that: $\angle R > \angle P$ and $\angle S > \angle Q$ .

Given:

$PQRS$ is a quadrilateral.

$PQ$ is its longest side and

$RS$ is its shortest side.
To prove:

$\angle R > \angle P$ and

$\angle S > \angle Q$
Construction: Join

$PR$ and

$QS$ .
Proof: In

$\Delta PQR$

$PQ$ is longest side (given)

$PQ > QR$

$\angle 5 > \angle 2$ …(i) (

$\angle$ opposite to longer side is greater)
In

$\Delta PSR$

$RS$ is the shortest side (given)

$PS > RS$

$\angle 6 > \angle 1$ …(ii)
Adding (i) and (ii), we get

$\angle 5 + \angle 6 > \angle 1 + \angle 2$

$\angle R > \angle P$
Now in

$\Delta PQS$ ,

$PQ$ is the longest side

$PQ > PS$

$\angle 8 > \angle 3$ …(iii)
In

$\Delta SRQ$ ,

$RS$ is the shortest side

$RQ > RS$ …(iv)

$\angle 7 > \angle 4$
Adding (iii) and (iv), we get

$\angle 7 + \angle 8 > \angle 3 + \angle 4$

$\angle S > \angle Q$
Hence proved.

In figure, side $QR$ of $\Delta PQR$ is produced on both sides such that $\angle PQS = \angle PRT$ . Prove that $PQ = PR$ .

$\Delta PQR$

$\angle PQS + \angle PQR = 180^{0}$ (linear pair angles)

$\angle PQS + \angle q = 180^{0}$

$\angle q = 180^{0} – \angle PQS$ …(i)
Also

$\angle r = 180^{0} – \angle PRT$ …(ii)
But

$\angle PQS = \angle PRT$ (given) …(iii)
From (iii), eqn (i) becomes

$\angle q = 180^{0} – \angle PRT$ …(iv)
Now from (ii) and (iv), we have

$\angle q = \angle r$
i.e.,

$\angle PQR = \angle PRQ$

$PR = PQ$ (converse of isosceles

$A$ theorem)
Hence proved.

If figure, $ABCD$ is a quadrilateral. Prove that $AB + BC + CD + DA > AC + BD$ .

$\Delta ABC$

$AB + BC > AC$ …(i) (the sum of any two sides of a

$\Delta$ is greater than third side)
Also in

$\Delta ADC$

$AD + DC > AC$ …(ii) (reason as above)
Similarly in

$\Delta ABD$ and

$\Delta BCD$

$AB + AD > BD$ …(iii)

$BC + CD > BD$ …(iv)
Adding (i), (ii), (iii) and (iv), we get

$2AB + 2BC + 2AD + 2CD > 2AC + 2BD$

$AB + BC + CD + DA > AC + BD$
Hence proved.

In figure, $T$ is a point on side $QR$ of $\Delta PQR$ and $S$ is a point such that $RT = ST$ . Prove that $PQ + PR > QS$ .

$\Delta PQR$ , we have

$PQ + PR > QR$ (the sum of two sides of a triangle is greater than third side).

$PQ + PR > QT + TR$

$[\because QT + TR = QR]$

$PQ + PR > QT + ST$ …(i)

$[\because TR = ST]$
In

$\Delta QST$ , we have

$QT + ST > QS$ …(ii)
From (i) and (ii), we get

$PQ + PR > QS$ .
Hence proved.

If two sides of a triangle are unequal then the longer side has greater angle opposite to it. Prove it.

Given: A

$\Delta ABC$ in which

$AC > AB$ (say)
To prove:

$\angle ABC > \angle ACB$
Construction: Mark a point

$D$ on

$AC$ such that

$AB = AD$ . Join

$BD$ .
Proof: In

$\Delta ABD$

$AB = AD$ (by construction)

$\angle 1 = \angle 2$ …(i) (angles opposite to equal sides are equal)
Now in

$\Delta BCD$

$\angle 2 > \angle DCB$ (ext. angle is greater than one of the opposite interior angles)

$\angle 2 > \angle ACB$ …(ii)

$[\because \angle ACB = \angle DCB]$
From (i) and (ii), we get

$\angle 1 > \angle ACB$ …(iii)
But

$\angle 1$ is a part of

$\angle ABC$

$\angle ABC > \angle 1$ …..(iv)
Now from (iii) and (iv), we get

$\angle ABC > \angle ACB$
Hence proved.
1870047_1878900_ans_6e8ac54ed4064ac99f10c76549cfa385.png

In figure, $PQR$ is a triangle and $S$ is any point in its interior, show that $SQ + SR < PQ + PR$ .

Given:

$S$ is any point in the interior of

$\Delta PQR$
To prove:

$SQ + SR < PQ + PR$
Construction: Produce

$QS$ to meet

$PR$ in

$T$ .
Proof: In

$\Delta PQT$ ,

$PQ + PT > QT$ (sum of two sides of

$\Delta$ is greater than third side)
or,

$PQ + PT > QS + ST$ …(i)
Also in

$\Delta RST$

$ST + TR > RS$ …(ii)
Adding (i) and (ii), we get

$PQ + PT + ST + TR > SQ + ST + SR$

$PQ + (PT + TR) > SQ + SR$

$PQ + PR > SQ + SR$
Hence proved.

For the following congruent triangles, find the pairs of corresponding angles.

To find: Pair of corresponding angles

Solution: In

$\Delta POQ$ and

$\Delta ROS$

$PO=RO$ (each

$4$ cm)

$OQ=OS$ (each

$4$ cm)

$PQ=RS$ (each

$5$ cm)

$\Delta POQ\cong \Delta ROS$ (By SSS cong. rule)

$\therefore$ corresponding

$\angle s$ are

$\angle P=\angle R$

$\angle POQ=\angle ROS$

$\angle Q=\angle S$ .

1190947_705141_ans_6629135616e348c9af8bf200d1de6893.jpg

In the adjoining figure, $AD=BD$ $\angle{ABD}={65}^{o}$ & $\angle{DAC}={22}^{o}$ .
Find $\angle{ACD}$

$AD=BD$

$\angle BAD=\angle ABD=65$

$\angle BAC=\angle BAD+\angle DAC$

$\angle BAC=65+22$

$\angle BAC=87$

$\angle A+\angle B+\angle C=180°$

$87+65+\angle C=180°$

$\angle C=180-152$

$\angle C=28$

$\therefore \angle ACD=28$

1128257_1038592_ans_1dc9eedd8c364c7c990771186cac485c.png

If $a, b, c$ are sides of a scalene triangle then prove that $(a + b + c)^{3} > 27$ $(a + b - c)(b + c - a)(c + a -b)$

$2s=a+b+c$

$\Rightarrow$

$a + b - c = 2s - 2c$

Similarly

$a-b+c=2s-2b,b+c-a=2s-2a$
(Using AM > GM, a

$\neq$ b

$\neq$ c )

$\displaystyle \frac{2s - 2c + 2s - 2a + 2s - 2b}{3} > [(a + b - c)(b + c - a)(c + a - b)]^{1/3}$

$\Rightarrow \displaystyle \frac{2s}{3} > [(a + b - c)(b + c - a)(c + a - b)]^{1/3}$

$\Rightarrow (a + b + c)> 3[(a + b - c)(b + c - a)(c + a - b)]^{1/3}$

$\Rightarrow (a + b + c)^{3} > 27(a + b - c)(b + c - a)(c + a - b)$

In given figures sides $AB$ and $BC$ and median $AD$ of a $\triangle ABC$ are respectively proportional to sides $PQ, QR$ and median $PM$ of $\triangle PQR$ . Show that $\triangle ABC \sim \triangle PQR$ .

Given

$AD$ and

$PM$ are medians of

$\triangle ABC$ and

$\triangle PQR,$

$\therefore$

$BD=\dfrac{1}{2}BC$ and

$QM=\dfrac{1}{2}QR........(1)$

Given that,

$\dfrac{AB}{PQ}=\dfrac{BC}{QR}=\dfrac{AD}{PM}.........(2)$

$\therefore$ From (1) and (2),

$\dfrac{AB}{PQ}=\dfrac{BD}{QM}=\dfrac{AD}{PM}...........(3)$

In

$\triangle ABD$ and

$\triangle PQM$

$\dfrac{AB}{PQ}=\dfrac{BD}{QM}=\dfrac{AD}{PM}$ [From(3)]

$\therefore$ By SSS criterian of proportionality

$\triangle ABD$

$\sim$

$\triangle PQM$

$\therefore$

$\angle B=\angle Q$ (Corresponding Sides of Similar Triangles)

$.............(4)$

In

$\triangle ABC$ and

$\triangle PQR$

$\dfrac{AB}{PQ}=\dfrac{BC}{QR}$ (From 2)

$\angle B=\angle Q$ (From 4)

$\therefore$ By SAS criterian of proportionality

$\triangle ABC$

$\sim$

$\triangle PQR$

$[hence \, proved]$

In given figure two sides $AB, BC$ and median $AM$ of one triangle $ABC$ are respectively equal to sides $PQ$ and $QR$ and median $PN$ of $\triangle PQR$ . Show that:
$\triangle ABM\cong \triangle PQN$

Given two sides

$AB, BC$ and median

$AM$ of

$\triangle ABC$ are respectively

equal to sides

$PQ$ and

$QR$ and median

$PN$ of

$\triangle PQR$ .

$BC=QR$ ..Given

$\therefore \cfrac 12 BC = \cfrac 12 QR$

$\therefore BM = QN$ ...(1)

$\triangle ABM$ and

$\triangle PQN$ ,

$AB=PQ$ ...Given

$AM=PN$ ...Given

$BM=QN$ ...From (1)

$\therefore \Delta ABM \cong \Delta PQN$ ...SSS test

$[hence \quad proved]$

In Fig $7.32$ , measure of some parts of triangles are given. By applying $RHS$ congruence rule, state which pairs of triangles are congruent. In case of congruent triangles, write the result in symbolic form.

Given,

to find out wether the given figures are

Congruent using

$RHS$

$(i)\$ Congruence rule

so, Now let us consider figure

$(i)$

as per

$RHS$ congruence rule.

$LQ=LF=90^o$ (

$\because$ given)

$PR=FD$ (given)

$PQ\neq DE$ (

$\because$ given in figure

$(i)$ )

$\therefore \ \triangle PQR$ is not congruence to

$\triangle DEF$

$(ii)$ Now let us consider figure

$(i)$

as per

$RHS$ congruence rule

$LC=LD=90^o$ (

$\because$ given )

$AC=BD$ (

$\because$ given)

$AB=AB$ (

$\because$ common side)

$\therefore \ \triangle ABC\cong \triangle ABD$

$(iii)$ Now let us consider figure

$(iii)$

as per

$RHS$ congruence rule we get,

$\angle B=\angle D=90^o$ (

$\because$ given)

$AB=AD$ (

$\because$ given)

$BC=CD$ (

$\because$ given)

$\therefore \ \triangle ABC\cong \triangle ADC$

$(iv)$ Let us consider figure

$(iv)$

as per

$RHS$ congurence rule we get,

$\angle S=\angle S=90^o$ (common angle )

$PQ=PR$ (

$\because$ given)

$QS=SR$ (

$\because$ common base)

$\therefore \ \triangle PSQ\cong \triangle PSR$

1417501_1088006_ans_c6665c7f19cc43e5b5793d6fb1174c07.JPG

If any point $O$ in the interior of a triangle ABC. Prove that AB $+$ BC $+$ CA $>$ OA $+$ OB $+$ OC.

$\begin{array}{l} In\, \Delta PBO \\ BP+PO>OB \\ Adding\, \, OC\, \, an\, both\, \, sides \\ we\, \, get, \\ BP+PO+OC>OB+OC\, \\ BP+PC>OB+OC \\ BP+PC>OB+OC\to (i)\, \, \, \left[ { PO+OC=PC } \right] \\ In\, \Delta APC \\ AP+AC>PC \\ Adding\, \, BP\, \, on\, \, both\, \, sides,\, we\, \, get \\ AP+AC+BP>PC+BP\to (ii)\, \, \, \left[ { AP+PB=AB } \right] \\ from\, \, \left( i \right) \, \, \& \, \, \left( { ii } \right) \\ AB+AC>BP+PC>OB+OC \\ AB+AC>OB+OC\to (iii) \\ similarly \\ BC+AC>OB+OA\to (iv) \\ AB+BC>OA+OC\to (v) \\ adding\, \, equation\, \, \left( 3 \right) \left( 4 \right) \left( 5 \right) \, \, we\, \, get \\ 2\left( { AB+BC+CA } \right) >2\left( { OA+OB+OC } \right) \\ AB+BC+CA>OA+OB+OC \end{array}$
1212068_1141319_ans_c9b2e28c92ca4d4290a06f3af2b71495.png

1212068_1141319_ans_c9b2e28c92ca4d4290a06f3af2b71495.png

$ABC$ is an isosceles triangle right angled at $B$ . Equilateral triangles $ACD$ and $ABE$ are constructed on sides $AC$ and $AB$ . Find the ratio between the areas of $\triangle{ABE}$ and $\triangle{ACD}$ .

It is given that $\Delta ABC$ is an isosceles triangle right angled at B, so,

$AB = BC$

In $\Delta ABC$ , by Pythagoras theorem,

$A{C^2} = A{B^2} + B{C^2}$

$A{C^2} = A{B^2} + A{B^2}$

$A{C^2} = 2A{B^2}$

As $\Delta ABE \sim \Delta ACD$ is given, then it is known that the ratio of areas of equilateral triangles is equal to the ratio of the squares of their corresponding sides, so,

$\dfrac{{{\rm{ar}}\left( {\Delta ABE} \right)}}{{{\rm{ar}}\left( {\Delta ACD} \right)}} = \dfrac{{A{B^2}}}{{A{C^2}}}$

$\dfrac{{{\rm{ar}}\left( {\Delta ABE} \right)}}{{{\rm{ar}}\left( {\Delta ACD} \right)}} = \dfrac{{A{B^2}}}{{2A{B^2}}}$

$\dfrac{{{\rm{ar}}\left( {\Delta ABE} \right)}}{{{\rm{ar}}\left( {\Delta ACD} \right)}} = \dfrac{1}{2}$

Therefore, ${\rm{ar}}\left( {\Delta ABE} \right):{\rm{ar}}\left( {\Delta ACD} \right) = 1:2$

In the adjacent figure $ABCD$ is a square and $\Delta A P B$ is an equilateral triangle. Prove that $\Delta A P D \cong \Delta B P C$ .
(Hint: In $\Delta A P D$ and $\Delta B P C \quad \overline { A D } = \overline { B C } , \overline { A P } = \overline { B P }$ and
$\angle P A D = \angle P B C = 90 ^ { \circ } - 60 ^ { \circ } = 30 ^ { \circ } )$

Given

$ABCD$ is square

$\therefore AB=BC=CD=AD$

$\angle A=\angle B=\angle C=\angle D=90^0$

Given

$PDB$ is equilateral

$\therefore AB=PD=PB$

$\angle (PAB)=\angle (APB)=\angle (PBA)=60^0$

$\therefore \angle (DAP)=\angle (DAB)-\angle (PAB)$

$=90-60=30^0$

$\therefore \angle(CBP)=\angle(CBA)-\angle(PBD)$

$=90-60=30^0$

$\Delta le$

$DPA$ and

$CPB$

$AD=BC$ [sides of the square]

$PA=PB$ [sides of the equilateral]

$\angle(DAP)=\angle(CBP)=60^0$

$\therefore \Delta DPA\equiv\Delta CPB$ [SAS axion]

$\Rightarrow \Delta DPA\cong \Delta BPC$

Hence proved.

1456066_1176688_ans_56efdd548b054ed58bf6cb05b289bb4f.png

O if any point in the interior of a triangle ABC. Prove that AB $+$ AC $>$ OB $+$ OC.

Construction=produced BO to meet AC in D

$\begin{array}{l} oof=In\, \Delta ABD \\ =AB+AD>BD \\ =AB+AD>BO+OD\to (i) \\ In\, \, \Delta OCD \\ CD+DO>OC\to (ii) \\ Adding\, \, equation\, \, (i)\, \, \& \, \, 9ii)\, \, we\, \, get \\ AB+AC>OB+OC\, \, \, \, \, \, \, \left[ { AD+DC=AC } \right] \end{array}$

1212335_1141318_ans_c62ae6902c2f4b07b4948eef765bfc36.PNG

Given : EB = DB
AE = BC
$\angle A\, = \,\angle C\, = \,90^\circ$
So, $\Delta ABE\, \cong \,\Delta CDB$

2041671_1079549_ans_d1040e5ed7354f33a26edd444dbde109.jpg

$ABC$ is an isosceles triangle in which $AC=BC\ AD$ and $BE$ are respectively two altitudes of side $BC$ and $AC$ . Prove that $AE=BD$ .

1170572_1125799_ans_94dcb5b867dd42d5b6ecf97d06b66a5b.jpg

In the given figure, $\angle \mathrm { BMN } = \angle \mathrm { CMN }$ and AN bisects $\angle B A C$ . Prove that $\Delta A B M \equiv \Delta A C M$ .

$\angle BMN=\angle CMN,AN$ bisects

$\angle BAC$

$\therefore$ In

$\triangle ABM$ and

$\triangle ACM$

$AM=AM$ (common)

$\angle MAB=\angle MAC$ (

$AN$ bisects

$\angle BAC$ )

Now

$\angle BMA=180^o-\angle BMN$

$\angle CMA=180^o-\angle CMN$

But

$\angle BMN=\angle CMN$

$\therefore \angle BMA=\angle CMA$

by ASA congruence rule

$\triangle ABM \cong\triangle ACM$

In the adjacent figure $ABCD$ is a square and $\Delta A P B$ is an equilateral triangle. Prove that $\Delta APD \cong \Delta PPC$
(Hint : In $\Delta APD$ and $\Delta B P C \quad \vec { A D } = \overline { B C } , \overline { A P } = \overline { B P }$ and $\angle P A D = \angle P B C = 90 ^ { \circ } - 60 ^ { \circ } = 30 ^ { \circ } ]$

Given Δ ABP is an equilateral triangle.

Then, AP = BP (Same side of the triangle)

and AD = BC (Same side of square)

and, ∠ DAP = ∠ DAB - ∠ PAB = 90° - 60° = 30°

Similarly,

∠ BPC = ∠ ABC - ∠ ABP = 90° - 60° = 30°

∴ ∠ DAP = ∠ BPC

∴ Δ APD is congruent to Δ BPC ( SAS proved)

In Δ APD

AP = AD II as AP = AB (equilateral triangle)

We know that ∠ DAP = 30°

∴ ∠ APD = (180° - 30°)/2 (Δ APD is an isosceles triangle and ∠ APD is on of the base angles.)

= 150°/2 = 75°

= ∠ APD = 75°

Similarly, ∠ BPC = 75°

Therefore, ∠ DPC = 360° - (75°+75°+60°)

= ∠ DPC = 150°

Now in Δ PDC

PD = PC as Δ APD is congruent to Δ BPC

∴ Δ PDC is an isosceles triangle

And ∠ PDC = ∠ PCD = (180° - 150°)/2

or ∠ PDC = ∠ PCD = 15°

∠ DPC = 150°; ∠ PDC = 15° and ∠ PCD = 15°

In the figure,prove that $\triangle APM \cong \triangle APN$ if $PM=PN, PM\bot AB$ and $PN \bot AC$

$PM+AB\left( { given } \right) \\ PN\bot AC\left( { given } \right) \\ \Delta ANP\, \, \, nad\, \, \, AMP \\ \angle PNA=90^{ \circ } \\ \angle PMA=90^{ \circ } \\ AP=AP\left( { common } \right) \\ Hence\, \, DM=PN \\ by\, \, S.A.S\, \, \, criteria \\$

In $\Delta PQR$ , If PQ>PR and bisectors of $\angle Q$ and $\angle R$ interested at $S$ . Show that SQ>SR.

Given that-

$\Delta PQR,PQ>PR$

$\angle Q\,\,\& \angle R$ have bisector intersected at

$S$

To find-

To prove

$SQ>SR$

$PQ>PR$

$\therefore \angle PQR> \angle PRQ$

$\therefore \dfrac{1}{2}\angle PQR> \dfrac{1}{2}\angle PRQ$

$\therefore \angle SRQ> \angle SQR$ [

$\because QS$ and

$RS$ bisectors of

$\angle Q$ and

$LR$ respectively]

$\Rightarrow SQ>SR$ (Since,

$\angle SRQ$ is larger and is formed by)

Hence, proved

$SQ$

1920598_1254204_ans_49e08793a448489b936d42c1bd86e501.png

Find the circumcentre of the triangle with the vertices A (-3,4) , B (3,4) and C (-4,-3). What is the circumradius of the circle?

We want the equation of circle to pass through points

$A\left( -3,4\right) , B\left( 3,4\right)$ and

$\left( -4,3\right)$

The equation of a circle is

$x^2+y^2+ax+by+c=0$

From the coordinates of the three given points we get three equations.

$\Rightarrow \left( -3\right)^2+4^2-3a+4b+c=0$

$\Rightarrow 3^2+4^2-3a+4b+c=0$

$\Rightarrow \left( -4\right)^2+3^2-4a+3b+c=0$

we can simplify the above equation -

$\Rightarrow \left[ \begin{matrix} -3 & 4 & 1 \\ 3 & 4 & 1 \\ -4 & 3 & 1 \end{matrix} \right] \left[ \begin{matrix} a \\ b \\ c \end{matrix} \right] =\left[ \begin{matrix} -25 \\ -25 \\ -25 \end{matrix} \right]$

solving the above matrix equation using craner's rule

$a=0,$

$b=0$ and

$c=-25$

Putting there values

$a,b$ and

$c$ in general form we write required equation of circle

$x^2+y^2-25=0$

On center radius form

$\left( x+0\right)^2+\left( y+0\right)^2=25$

Center of circle is

$\left( 0,0\right)$ and radius

$=5.$

Hence, the answer is

$5.$

$ABCD$ is a square. $E$ and $F$ are respectively the mid-point of $BC$ and $CD$ . If $R$ is the mid point of $EF$ prove that
$ar(AER)=ar(AFR)$ .

GivenL

$ABCD$ is a square.

$E$ and

$F$ are the midpoints of

$BC$ and

$CD$ and

$R$ is the midpoint of

$EF$

Proof:Consider

$\triangle{ABE}$ and

$\triangle{ADF}$

We know that

$BC=DC$ (all sides of a square are equal.)

$\Rightarrow \dfrac{BC}{2}=\dfrac{DC}{2}$

$\Rightarrow EB=DF$ (since

$E$ and

$F$ are midpoints of

$BC$ and

$CD$ )

$\angle{ABE}=\angle{ADF}={90}^{\circ}$ (All angles in a square are right angles)

$AB=AD$ (All sides in a square are equal)

$\therefore$ by SAS criterion,

$\triangle{ABE}\cong\triangle{ADF}$

Hence,

$AE=AF$ by C.P.C.T

$\Rightarrow$ ,

$R$ is the midpoint of

$EF$

$AR$ is the median of the triangle

$AEF$

Since median divides the triangle into two triangles of equal areas,ar

$\left(\triangle{AER}\right)=$ ar

$\left(\triangle{AFR}\right)$

Hence proved.

In right triangle $ABC$ , right angled at $C,M$ is the mid-point of hypotenuse $AB$ . $C$ is joined to $M$ and produced to a point $D$ such that $DM = CM$ . Point $D$ is joined to point $B$ . Show that $CM = \dfrac{1}{2}AB$

$AM=BM=\dfrac{1}{2}AB$

$\Rightarrow \angle =90°$

$DM=CM$

$\triangle DBM$ and

$\triangle BCM$

$\Rightarrow BM=BM$ ( common )

$DM=CM$ ( given )

$\angle DMB\cong \triangle BCM$

$\therefore BM=CM=\dfrac{1}{2}AB$

Hence, solved.

1227898_1343468_ans_7fb4252701464946832d9c8d629c7cc1.png

In figure ,if $seg\,PR \cong \,seg\,PQ,$
show that $seg\,PS > \,seg\,PQ.$

1961036_1438463_ans_e750cc7095f641ada80bb5d676c57cca.jpg

If line segments AB and CD bisect each other
at O, then using SAS congruence rule prove
that $\Delta AOC \equiv \Delta BOD$ .

2040892_1453273_ans_66182f4a848045c4b20fb54ae83adcab.jpg

Show that the origin in with in the triangle whose angular points are (2, 1), (3, -2) and (-4, -1).

2045319_1570207_ans_28498cce62844172aa58970d239221bb.jpg

In the following diagram, AP and BQ are equal and parallel to each other.
Prove that:
(i) $\Delta AOP \equiv \Delta BOQ$
(ii) $AB$ and $PQ$ bisect each other

2045447_1572431_ans_7db3038a8d80443598c0d0eb5f1420db.jpg

In a $\Delta PQR$ , if $PQ=QR$ and L, M and N are mid-points of the sides PQ,QR and RP respectively. Prove that $LN=MN$ .

Hence, LN=MN
1653338_1669566_ans_c60abdb62f7f40cdb2d6d95b5ff6a15b.jpg

In right triangle $ABC$ , right angled at $C$ , $M$ is the mid-point of hypotenuse $AB$ . $C$ is joined to $M$ and produced to a point $D$ such that $DM = CM$ . Point $D$ is joined to point $B$ (see the given figure). Show that: $(i) \triangle{AMC}\cong \triangle{BMD}$ $\left(ii\right) CM = \dfrac{1}{2}AB$

Proof (i) In

$\Delta AMC$ and

$\Delta BMD$

M is the midpoint

$\Rightarrow AM = BM$

$\Rightarrow \angle CMA = \angle DMB$ (vertically opposite angles)

$CM = DM$ (given)

Hence

$\Delta AMC \cong \Delta BMD$ by

$SAS$ congruence rule

(ii)

$DC = AB$ Since

$\Delta DBC \cong \Delta ACB$

$\Rightarrow DM+CM = AB$

$\Rightarrow CM+CM = AB$ Since

$CD = CM+DM$

$\Rightarrow 2CM = AB$ and

$CM = DM$

Hence

$CM = 1/2 AB$

1240665_1505783_ans_fcd931e0da1b4c2a844ddec1e0f95fd7.PNG

Can you construct a triangle ABC with BC $=6$ cm, $\angle$ B $=60^o$ and AB $+$ AC $=5$ cm? If not give reason.

1319293_1585045_ans_049dea1cab094020bb1028e2e1114956.jpg

$ABC$ and $D B C$ are two isosceles triangles on the same base $B C$ (see Fig.). Show that $\angle ABD=\angle ACD.$

Given :-

$\Delta ABC$ is isosceles

$AB = AC$ __(1)

$\Delta DBC$ is isosceles

$DB = DC$ __(2)

To prove :-

$\angle ABD = \angle ACD$

Proof :- let us join

$AD$

$\Delta ABD$ and

$\Delta ACD$ ,

$AB = AC$ [from eq. (1)]

$BD = CD$ [ from eq. (2)]

$AD = AD$ [common]

$\Delta ABD \cong \Delta ACD$ (sss rule)

$\angle ABD = \angle ACD$ (CPCT)

Hence proved

1233745_1502059_ans_6deedf23386c4bc581566ffc398549f6.png

If $\Delta ABC \sim \Delta PQR, \ ar( \Delta ABC)=16 \ cm^2$ and $\ ar(\Delta PQR)=81 \ cm^2, AB=2 \ cm$ find PQ.

2043830_1538106_ans_b561895d0d024b4197badfc2a85130ec.jpg

In $\Delta ABC,BD \perp AC$ and $CE \perp AB.$ If BD and CE intersect at O, prove that $\angle BOC = 180^{\circ} - A.$

1869729_1669822_ans_64f52b34ef414f1b8960fa5ae6b0fd7d.png

In a $\Delta ABC, AD$ bisects $\angle A$ and $\angle C > \angle B.$ Prove that $\angle ADB > \angle ADC.$

1653789_1669821_ans_e2bacd763eb2431c8712e5a6f44e335f.jpg

In figure, P is the point on the side BC. Complete the following statement using symbol $' = ' , ' > '$ or $' < '$ so as to make it true.
$AP.....AB+BP$ .

2039869_1677868_ans_1b047133d1a54b80b9276d8858eb6eee.png

Triangles - Class 9 Maths - Extra Questions

In △ABC\triangle ABC and △PQR\triangle PQR, BC=QR,∠A=90∘,∠C=∠R=40∘ and ∠Q=50∘.BC=QR, \angle A=90^{\circ},\angle C =\angle R=40^{\circ}\ and\ \angle Q=50^{\circ}.Above two triangles are congruent by ASA testIf true then enter 11 else if False enter 0.0.

In △ABC and△PQR,AB=PQ,AC=PRandBC=QR\bigtriangleup ABC \ and \bigtriangleup PQR, AB=PQ,AC=PR\:and\:BC =QR, then the two triangles are congruent by which test?

In Δ\DeltaPQR, seg PQPQ ≅\cong seg QRQR and ∠P=70o\angle P = 70^o, then find the measures of angle QQ is:

Without drawing exact triangles, state whether the given pairs of triangle are congruent or not, if congruent use '1' else '0'.In ΔABCandΔDEF;AB=EF,BC=DFand∠B=∠F \Delta ABC \,\,\, and \,\,\, \Delta DEF; AB=EF, BC=DF \,\, and \,\, \angle B = \angle F

In ΔABC,\Delta ABC, if AB=76 AB = 76 cm, BC=69BC=69 cm and CA=61CA=61 cm, then the smallest angle is BB.If true then enter 11 and if false then enter 00.

In the alongside diagram, ABCDABCD is parallelogram. Prove that : AB=2BCAB = 2 BC

Without drawing exact triangles, state, giving reasons, whether the given pairs of triangle are congruent or not :ΔPQRandΔSQR;PQ=SRand∠PQR=∠SQR \Delta PQR \,\,\, and \,\,\, \Delta SQR; PQ=SR\,\, and \,\, \angle PQR = \angle SQR.

Without drawing exact triangles, state, giving reasons, whether the given pairs of triangle are congruent or not :In ΔABC and ΔPQR;AB=QR,AC=PRand∠B=∠R\Delta ABC \ and \ \Delta PQR; AB=QR, AC=PR \,\, and \,\, \angle B = \angle R.

If the two sides and the included angle of a triangle are respectively equal to two sides and the included angle of the other triangle, the two triangles are congruent. If true enter 1 else 0.

If △DEF≅△RPQ\triangle DEF\cong \triangle RPQ, then ∠D=∠Q\angle D=\angle Q.If true enter 11, else 00.

If △PQR≅△CAB\triangle PQR\cong \triangle CAB, PQ=CAPQ = CA.If true enter 11, else 00.

If △ABC≅△PRQ\triangle ABC\cong \triangle PRQ then AB=PQAB = PQ.If true enter 11, else 00.

State whether the given statement is True(Press 1) or False(Press 0):If the two angles and the included sides of one triangle are respectively equal to the two angles and the included side of the other triangle, then the two triangles are congruent.

Two right triangles are congruent, if hypotenuse and a side of one triangle are respectively equal to the hypotenuse and a side of the other triangle. If true enter 1 else 0.

State true or false:If the two sides of a triangle are unequal, then the larger side has the smaller angle opposite to it.If true enter 1, else 0.

Prove that the circle drawn with any equal side of an isosceles triangle as diameter bisects the base

AB is a line segment AX and BY are two equal line segments drawn opposite sides of line AB such that AX ||BY. If line segments AB and XY intersect each other at point P. Prove that (a) ΔAPX≅ΔBPY\Delta APX \cong \Delta BPY (b) line segments AB and XY bisect each other at P.

In ΔABC\Delta ABC, the bisector AX of ∠A\angle A intersects BC at X. XL⊥ABXL\perp AB and XM⊥ACXM \perp AC are drawn. Is XL = XM?Why or why not?

Two parallel lines ll and mm are intersected by another pair of parallel lines pp and qq as in the figure. Show that triangles ABCABC and CDACDA are congruent.

In the figure AD = BC and BD = CA. Prove that ∠ADB=∠BCA\angle ADB = \angle BCA and ∠DAB=∠CBA\angle DAB = \angle CBA.

In △ABC,AB=BC\triangle ABC, AB = BC and ∠B=64o\angle B = 64^o. Find ∠C\angle C.

In the figure, C is the mid point of AB, ∠BAD=∠CBE,∠ECA=∠DCB\angle BAD = \angle CBE, \angle ECA= \angle DCB. Prove that (a) ΔDAC≅ΔEBC\Delta DAC \cong \Delta EBC (b) DA = EB.

In a quadrilateral ACBC, AC = AD and AB bisect ∠A\angle A. Show that △ABC\triangle ABC is congruent to △ABD\triangle ABD.

In the figure, it is given that AE = AD and BD = CE. Prove that △AEB\triangle AEB is congruent △ADC\triangle ADC.

In △ABC,AD\triangle ABC, AD is the perpendicular bisector of BCBC (see given figure). Show that △ABC\triangle ABC is an isosceles triangle in which AB=ACAB = AC

In the given figure, the point PP bisects ABAB and DCDC. Prove that△APC≅△BPD\triangle APC\cong \triangle BPD

The Indian Navy flight fly in a formation that can be viewed as two triangles with common side. Prove that △SRT≅△QRT\triangle SRT \cong \triangle QRT, if TT is the midpoint of SQSQ and SR=RQSR = RQ.

ABCABC is an isosceles triangle in which altitudes BDBD and CECE are drawn to equal sides ACAC and ABAB respectively (see figure). Show that these altitudes are equal

In the figure given below, AB=DCAB = DC and AC=DBAC = DB. Is ΔABC≅ΔDCB\Delta ABC \cong \Delta DCB.

Complete the congruence statement.ΔABC≡\Delta ABC \equiv ?

DE = EC show that AB + BC > AD.

If SS is any point in the interior of ΔPQR\Delta PQR, prove that (SQ+SR)<(PQ+PR)(SQ + SR) < (PQ + PR).

Prove that a median divides a triangle into equal parts

In the adjoining figure ∠ABD=132o\angle{ABD}={132}^{o} & ∠EAC=120o\angle{EAC}={120}^{o}. Prove that AB>ACAB>AC

Write down the total number of True statement:Take any point O in the interior of a triangle PQR.Is(i) OP + OQ > PQ?(ii) OQ + OR > QR?(iii) OR + OP > RP?

"If two sides and an angle of one triangle are equal to two sides and an angle of another triangle, then the two triangles must be congruent". Is the statement true? Why?

In figure The sides BABA and CACA have been produced such that BA=ADBA=AD and CA=AECA=AE prove that segment DE∥BCDE\parallel BC

In triangles ABCABC and PQR,∠A=∠QPQR, \angle A = \angle Q and ∠B=∠R\angle B = \angle R. Which side of ΔPQR\Delta PQR should be equal to side ABAB of ΔABC\Delta ABC so that the two triangle are congruent? Give reason for your answer.

In fig., PQRS is square and SRT is an equilateral triangle. Prove that(i) PT=QT PT = QT (ii) ∠TQR=15∘\angle TQR = {15^ \circ }

Give any two real-life examples for congruent shapes.

In figure, PR>POPR>PO and PSPS bisect OPROPR. Prove that ∠PSR>∠PSO\angle{PSR}>\angle{PSO}

If the point p(a2,a)p\left( {{a^2},\,a} \right)\, lies in the region. Corresponding in the acute angle between the lines 2y=xand4y=x,2y = x\,\,and\,\,4y = x, then find the value of aa or the range in which aa lies.

In the given fig △PQC\triangle PQC & PRCPRC such that QC=CRQC=CR, PQ=PRPQ=PR. Prove that △PQC≅△CPR\triangle PQC\cong\triangle CPR

If the area of two similar triangles are equal, prove that they are congruent.

Give any two real-life examples of congruent shapes.

SS and TT are respectively the mid points of equal sides PQPQ and PRPR of △PQR\triangle {PQR}. Show that Qt=RSQt=RS.

You want to show that ΔART≅ΔPEN\Delta ART \cong \Delta PEN.If it is given that AT=PNAT = PN and you are to use ASA criterion, you need to have i) ?ii) ?

Δ\DeltaABC and Δ\DeltaDBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC. If AD is extended to intersect BC at P, show that Δ\Delta ABP≅\cong Δ\Delta ACP.

In △PQR,PQ=PR\triangle PQR, PQ=PR and m∠P=40om \angle P=40^{o} then m∠Q=.m \angle Q=.

In a △ABC,BC=a,CA=b\triangle ABC,BC=a,CA=b and ∠BCA=120o,CD\angle BCA=120^{o},CD is angle bisector of ∠BCA\angle BCA which meets ABAB on DD find the length of CDCD

In the given figure ΔABC≅ΔABT\Delta ABC \cong \, \Delta ABT, write all the corresponding sides.

In the adjoining figure, PQ==PR and QL==MR. Prove that PL==PM.

Explain, why ?ΔABC=ΔFED\Delta A B C = \Delta F E D

In figure ABCDABCD, ABFEABFE and CDEFCDEF are parallelograms. Prove that ar(△ADE)=ar(△BCF)ar(\triangle {ADE})=ar(\triangle{BCF)}

If △ABC≅△FED\bigtriangleup ABC\cong\bigtriangleup FED under the corresponding ABC↔FED. under\ the\ corresponding \ ABC \leftrightarrow FED. Write all the corresponding congruent part of the triangle.

In the following figure, ∠A=∠C\angle A = \angle C and AB=BC.AB = BC.Prove that ΔABD≅ΔCBE\Delta ABD \cong \Delta CBE

Prove that in a right angled triangle, hypotenuse is the longest side.

Given: ΔA⊥AB\Delta A \bot AB, CB⊥ABCB \bot AB and AC=BDAC = BDState 33 points of equal pairs in ΔABC\Delta ABC & ΔABD\Delta ABD & Congruence rule.

ABC is a right isosceles triangle right angled at A. Find the value of ∠B\angle B and ∠C\angle C.

In the given figure, equal sides BABA and CACA of ΔABC\Delta ABC are produced to QQ and PP respectively such that AP=AQAP = AQ. Prove that PB=QCPB = QC.

PA is tangent to the circle with centre O. If BC = 3cm, AC = 4 cm and ΔACB∼ΔPAO,\Delta ACB \sim \Delta PAO, then find OA and OPAP\frac{OP}{AP}.

Is it possible to have a triangle with the following sides?2 cm, 3 cm,5 cm

In figure, equal sides have been marked by the same signs.Can we say ΔABC≅ΔDBC\Delta ABC\cong \Delta DBC?

In △ABC\triangle ABC, AD is perpendicular bisector of BC (See adjacent figure). Show that △ABC\triangle ABC is an isosceles triangle in which AB=ACAB=AC

In the figure, AP=AQAP =AQ and BP=BQ.BP= BQ.Prove that ABAB is the bisector of ∠PAQ\angle PAQ and ∠PBQ.\angle PBQ.

In figure, AB=ACAB=AC and ∠1=∠2\angle 1=\angle 2. Is ΔABD≅ΔACD\Delta ABD\cong \Delta ACD ?(Give reason)

If ΔABC≅ΔFED\Delta A B C \cong \Delta F E D under the correspondence ABC↔FEDA B C \leftrightarrow F E D, write all the corresponding congruent parts of the triangles.

If ΔABC≅△NMO,\Delta A B C \cong \triangle N M O , name the congruent sides and angles.

Consider the figure, PQRSPQRS is a rectangle. PTPT and RURU are perpendicular from PP and RR on SQSQ. Prove that ΔPTS≅ΔQRU\Delta PTS\cong \Delta QRU

In figure, equal sides have been marked by the same signs.Is ΔABC≅ΔDCB\Delta ABC\cong \Delta DCB?

In the figure, ΔCDE\Delta CDE is an equilateral triangle formed on a side CDCD of a square ABCDABCD. Show that ΔADE≅ΔBCE\Delta ADE \cong \Delta BCE.

In the given figure, prove that: AC=DBAC=DB

One of the angles of a △\triangle is 75.{ 75 }^{ }. If the difference of the other two other is 35.{ 35 }^{ }.Find the largest angle of the △.\triangle .

In △ABC\triangle{ABC},AB=ACAB=AC and ADAD is the perpendicular bisector of BCBC. Show that △ADB≅△ADC\triangle{ADB}\cong\triangle{ADC}

In the given figure, prove that: △ABC≅△DCB\triangle ABC\cong \triangle DCB

If the area of two similar triangles are equal, prove that they are congruent.

I the given figure, we have PQ=SRPQ=SR and PR=SQPR=SQ. Prove that △PQR≅△SRQ\triangle PQR \cong \triangle SRQ.

In the givenABAB and CDCD bisect each other at OO.State the three pairs of equal parts in two triangles AOCAOC and BODBOD

If ΔPQR\Delta PQR is an isosceles triangle such that PQ=PR , then prove that the attitude PS from P on QR bisects QR

In $\triangle ABC$ and $\triangle PQR$ , $BC=QR, \angle A=90^{\circ},\angle C =\angle R=40^{\circ}\ and\ \angle Q=50^{\circ}.$
Above two triangles are congruent by ASA test
If true then enter $1$ else if False enter $0.$

In $\bigtriangleup ABC \ and \bigtriangleup PQR, AB=PQ,AC=PR\:and\:BC =QR$ , then the two triangles are congruent by which test?

In $\Delta$ PQR, seg $PQ$ $\cong$ seg $QR$ and $\angle P = 70^o$ , then find the measures of angle $Q$ is:

Without drawing exact triangles, state whether the given pairs of triangle are congruent or not, if congruent use '1' else '0'.
In $\Delta ABC \,\,\, and \,\,\, \Delta DEF; AB=EF, BC=DF \,\, and \,\, \angle B = \angle F$

In $\Delta ABC,$ if $AB = 76$ cm, $BC=69$ cm and $CA=61$ cm, then the smallest angle is $B$ .
If true then enter $1$ and if false then enter $0$ .

In the alongside diagram, $ABCD$ is parallelogram. Prove that : $AB = 2 BC$

Without drawing exact triangles, state, giving reasons, whether the given pairs of triangle are congruent or not :
$\Delta PQR \,\,\, and \,\,\, \Delta SQR; PQ=SR\,\, and \,\, \angle PQR = \angle SQR$ .

Without drawing exact triangles, state, giving reasons, whether the given pairs of triangle are congruent or not :
In $\Delta ABC \ and \ \Delta PQR; AB=QR, AC=PR \,\, and \,\, \angle B = \angle R$ .

If $\triangle DEF\cong \triangle RPQ$ , then $\angle D=\angle Q$ .
If true enter $1$ , else $0$ .

If $\triangle PQR\cong \triangle CAB$ , $PQ = CA$ .
If true enter $1$ , else $0$ .

If $\triangle ABC\cong \triangle PRQ$ then $AB = PQ$ .
If true enter $1$ , else $0$ .

State whether the given statement is True(Press 1) or False(Press 0):
If the two angles and the included sides of one triangle are respectively equal to the two angles and the included side of the other triangle, then the two triangles are congruent.

State true or false:
If the two sides of a triangle are unequal, then the larger side has the smaller angle opposite to it.
If true enter 1, else 0.

AB is a line segment AX and BY are two equal line segments drawn opposite sides of line AB such that AX ||BY.
If line segments AB and XY intersect each other at point P. Prove that
(a) $\Delta APX \cong \Delta BPY$ (b) line segments AB and XY bisect each other at P.

In $\Delta ABC$ , the bisector AX of $\angle A$ intersects BC at X. $XL\perp AB$ and $XM \perp AC$ are drawn. Is XL = XM?
Why or why not?

Two parallel lines $l$ and $m$ are intersected by another pair of parallel lines $p$ and $q$ as in the figure. Show that triangles $ABC$ and $CDA$ are congruent.

In the figure AD = BC and BD = CA. Prove that $\angle ADB = \angle BCA$ and $\angle DAB = \angle CBA$ .

In $\triangle ABC, AB = BC$ and $\angle B = 64^o$ . Find $\angle C$ .

In the figure, C is the mid point of AB, $\angle BAD = \angle CBE, \angle ECA= \angle DCB$ .
Prove that (a) $\Delta DAC \cong \Delta EBC$ (b) DA = EB.

In a quadrilateral ACBC, AC = AD and AB bisect $\angle A$ . Show that $\triangle ABC$ is congruent to $\triangle ABD$ .

In the figure, it is given that AE = AD and BD = CE. Prove that $\triangle AEB$ is congruent $\triangle ADC$ .

In $\triangle ABC, AD$ is the perpendicular bisector of $BC$ (see given figure). Show that $\triangle ABC$ is an isosceles triangle in which $AB = AC$

In the given figure, the point $P$ bisects $AB$ and $DC$ . Prove that
$\triangle APC\cong \triangle BPD$

The Indian Navy flight fly in a formation that can be viewed as two triangles with common side. Prove that $\triangle SRT \cong \triangle QRT$ , if $T$ is the midpoint of $SQ$ and $SR = RQ$ .

$ABC$ is an isosceles triangle in which altitudes $BD$ and $CE$ are drawn to equal sides $AC$ and $AB$ respectively (see figure). Show that these altitudes are equal

In the figure given below, $AB = DC$ and $AC = DB$ . Is $\Delta ABC \cong \Delta DCB$ .

Complete the congruence statement.
$\Delta ABC \equiv$ ?

If $S$ is any point in the interior of $\Delta PQR$ , prove that $(SQ + SR) < (PQ + PR)$ .

In the adjoining figure $\angle{ABD}={132}^{o}$ & $\angle{EAC}={120}^{o}$ . Prove that $AB>AC$

Write down the total number of True statement:
Take any point O in the interior of a triangle PQR.
Is
(i) OP + OQ > PQ?
(ii) OQ + OR > QR?
(iii) OR + OP > RP?

In figure The sides $BA$ and $CA$ have been produced such that $BA=AD$ and $CA=AE$ prove that segment $DE\parallel BC$

In triangles $ABC$ and $PQR, \angle A = \angle Q$ and $\angle B = \angle R$ . Which side of $\Delta PQR$ should be equal to side $AB$ of $\Delta ABC$ so that the two triangle are congruent? Give reason for your answer.

In fig., PQRS is square and SRT is an equilateral triangle. Prove that
(i) $PT = QT$ (ii) $\angle TQR = {15^ \circ }$

In figure, $PR>PO$ and $PS$ bisect $OPR$ . Prove that $\angle{PSR}>\angle{PSO}$

If the point $p\left( {{a^2},\,a} \right)\,$ lies in the region. Corresponding in the acute angle between the lines $2y = x\,\,and\,\,4y = x,$ then find the value of $a$ or the range in which $a$ lies.

In the given fig $\triangle PQC$ & $PRC$ such that $QC=CR$ , $PQ=PR$ . Prove that $\triangle PQC\cong\triangle CPR$

$S$ and $T$ are respectively the mid points of equal sides $PQ$ and $PR$ of $\triangle {PQR}$ . Show that $Qt=RS$ .

You want to show that $\Delta ART \cong \Delta PEN$ .
If it is given that $AT = PN$ and you are to use ASA criterion, you need to have
i) ?
ii) ?

$\Delta$ ABC and $\Delta$ DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC. If AD is extended to intersect BC at P, show that $\Delta$ ABP $\cong$ $\Delta$ ACP.

In $\triangle PQR, PQ=PR$ and $m \angle P=40^{o}$ then $m \angle Q=.$

In a $\triangle ABC,BC=a,CA=b$ and $\angle BCA=120^{o},CD$ is angle bisector of $\angle BCA$ which meets $AB$ on $D$ find the length of $CD$

In the given figure $\Delta ABC \cong \, \Delta ABT$ , write all the corresponding sides.

In the adjoining figure, PQ $=$ PR and QL $=$ MR. Prove that PL $=$ PM.

Explain, why ?
$\Delta A B C = \Delta F E D$

In figure $ABCD$ , $ABFE$ and $CDEF$ are parallelograms. Prove that $ar(\triangle {ADE})=ar(\triangle{BCF)}$

If $\bigtriangleup ABC\cong\bigtriangleup FED$ $under\ the\ corresponding \ ABC \leftrightarrow FED.$ Write all the corresponding congruent part of the triangle.

In the following figure, $\angle A = \angle C$ and $AB = BC.$
Prove that $\Delta ABD \cong \Delta CBE$

Given: $\Delta A \bot AB$ , $CB \bot AB$ and $AC = BD$
State $3$ points of equal pairs in $\Delta ABC$ & $\Delta ABD$ & Congruence rule.

ABC is a right isosceles triangle right angled at A. Find the value of $\angle B$ and $\angle C$ .

In the given figure, equal sides $BA$ and $CA$ of $\Delta ABC$ are produced to $Q$ and $P$ respectively such that $AP = AQ$ . Prove that $PB = QC$ .

PA is tangent to the circle with centre O. If BC = 3cm, AC = 4 cm and $\Delta ACB \sim \Delta PAO,$ then find OA and $\frac{OP}{AP}$ .

Is it possible to have a triangle with the following sides?
2 cm, 3 cm,5 cm

In figure, equal sides have been marked by the same signs.
Can we say $\Delta ABC\cong \Delta DBC$ ?

In $\triangle ABC$ , AD is perpendicular bisector of BC (See adjacent figure). Show that $\triangle ABC$ is an isosceles triangle in which $AB=AC$

In the figure, $AP =AQ$ and $BP= BQ.$

Prove that $AB$ is the bisector of $\angle PAQ$ and $\angle PBQ.$

In figure, $AB=AC$ and $\angle 1=\angle 2$ . Is $\Delta ABD\cong \Delta ACD$ ?(Give reason)

If $\Delta A B C \cong \Delta F E D$ under the correspondence $A B C \leftrightarrow F E D$ , write all the corresponding congruent parts of the triangles.

If $\Delta A B C \cong \triangle N M O ,$ name the congruent sides and angles.

Consider the figure, $PQRS$ is a rectangle. $PT$ and $RU$ are perpendicular from $P$ and $R$ on $SQ$ . Prove that $\Delta PTS\cong \Delta QRU$

In figure, equal sides have been marked by the same signs.
Is $\Delta ABC\cong \Delta DCB$ ?

In the figure, $\Delta CDE$ is an equilateral triangle formed on a side $CD$ of a square $ABCD$ . Show that $\Delta ADE \cong \Delta BCE$ .

In the given figure, prove that:
$AC=DB$

One of the angles of a $\triangle$ is ${ 75 }^{ }.$ If the difference of the other two other is ${ 35 }^{ }.$
Find the largest angle of the $\triangle .$

In $\triangle{ABC}$ , $AB=AC$ and $AD$ is the perpendicular bisector of $BC$ . Show that $\triangle{ADB}\cong\triangle{ADC}$

In the given figure, prove that:
$\triangle ABC\cong \triangle DCB$

I the given figure, we have $PQ=SR$ and $PR=SQ$ .
Prove that $\triangle PQR \cong \triangle SRQ$ .

In the given $AB$ and $CD$ bisect each other at $O$ .State the three pairs of equal parts in two triangles $AOC$ and $BOD$

If $\Delta PQR$ is an isosceles triangle such that PQ=PR , then prove that the attitude PS from P on QR bisects QR

You want to establish $\triangle DEF\cong \triangle MNP$ , using ASA congruence rule. You are given that $\angle D=\angle M$ and $\angle F=\angle P$ . What information is needed to establish the congruence ?

Check whether $\triangle{ABC}\cong\triangle{DEF}$ .Give Reason
In $\triangle ABC: AB=6\ cm,\angle B=50^{\circ}$ and $\angle C=90^{\circ}$
In $\triangle DEF: DE=6\ cm,\angle E=50^{\circ}, \angle F=90^{\circ}$ .

In the given figure, $\overline{PL}\bot \overline{OB}$ and $\overline{PM}\bot \overline{OA}$ such that $\overline{PL}=\overline{PM}$ . Is $\triangle PLO\cong \triangle PMO$ ? Give reasons in support of your answer.

Write converse of the theorem "In $\triangle ABC$ , if $AB=AC$ then $\angle C=\angle B$ "

In the given figure, PS is the bisector of $\angle QPR$ of $\Delta PQR$ Prove that $\frac{QS}{SR}=\frac{PQ}{PR}$

Take any point $O$ in the interior of a $\triangle{PQR}$ .Is $OR+OP>RP$ ?

Take any point O in the interior of a triangle PQR. is
$OQ+OR>QR?$

Is $\Delta ABC\cong \Delta DEF?$ Give reasons in support of your answer.

Without drawing the triangles write all size pairs of equal measures in each of the following pairs of congruent triangles.
$\triangle ABC \cong \triangle LMN$

In $\triangle PQR$ and $\triangle SQR$ are both isosceles triangle on a common base $QR$ such that $P$ and $S$ lie on the same side of $QR$ . Are triangles $PSQ$ and $PSR$ congruent ? Which condition do you use ?

In Fig.6.49, it is given that $LM = ON$ and $NL = MO$
(a) State the three pairs of equal parts in the triangles $NOM$ and $MLN$ .
(b) Is $\triangle NOM \cong \triangle MLN$ . Give reason ?

Without drawing the triangles write all size pairs of equal measures in each of the following pairs of congruent triangles.
$\triangle STU \cong \triangle DEF$