Explanation
Let the bisector of $$∠A$$ bisects the side $$BC$$ at $$X$$.
Given, $$ABCD$$ is a parallelogram.
∴ $$AD||BC$$ (Opposite sides of the parallelogram are parallel)
Now, $$AD||BC$$ and AX is the transversal ,
∴ $$∠2 = ∠3$$ (Alternate angles) ............(1)
and $$∠1 = ∠2$$ ($$AX$$ is the bisector of $$∠A$$) ................(2)
From (1) and (2), we obtain
$$∠1 = ∠3$$
Now, in $$ΔABX$$,
⇒ $$AB = BX$$ ( If two angles of a triangle are equal, then sides opposite to
them are equal)
⇒ $$2AB = 2BX = BX + BX = BX + XC$$ ( $$X$$ is the mid point of $$BC$$)
⇒ $$2AB = BC$$
⇒ $$2AB = AD$$ (Opposite sides of a parallelogram are equal)
∴ $$AD = 2AB.$$
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