Explanation
In $$ \triangle ABD $$, we have sum of two sides $$>$$ third side $$ => AB + AD > BD $$ OR
$$ => AB + AD > BE + ED -- (1) $$Also, in $$ \triangle BCE $$, we have sum of two sides $$>$$ third side $$ => EB + EC > BC $$ ---- (2)Adding equations $$ 1, 2 $$ we get$$ => AB + AD + EB +EC > BE + ED +BC $$$$ => AB + AD + EC > ED + BC $$
But $$ EC = ED $$
Hence, $$ AB + AD > BC$$
by isosceles triangle property, angles opposite to equal sides are equal.
Then, let $$\angle CAD = \angle CDA = x$$.
In a $$\triangle$$ $$PQR, QR = PR$$ and $$\angle\,P\,=\,36^{\circ}.$$ The largest side of the triangle is:
Since, $$ QR = PR $$, their opposite angles are also equal.Thus $$ \angle P = \angle Q = {36}^{0} $$ Since sum of angles in a triangle PQR $$ = {180}^{o} $$$$ \Rightarrow \angle R = 180 - 2(36) = {108}^{o} $$
Thus, $$ \angle R $$ is the largest angle of the triangle, the side opposite to it, which is $$PQ$$ will be the greatest side of the triangle.
By which congruency are the following pair of triangles congruent:
In $$\Delta\,ABC$$ and $$\Delta \,DEF$$, $$\angle\,B = \angle\,E = 90\,^{\circ}, AC = DF$$ and $$BC = EF.$$
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