Question

In a parallelogram ABCD, the bisector of angle A also bisects BC at x. Prove that AD=2AB

Answer 1

In parallelogram ABCD ,

Bisector of ∠A bisects BC at X

∵ AD││BC and AX cuts them so

∠DAX = ∠AXB (alternate angles)

∠DAX = ∠XAB (AX is bisector of ∠A)

∴∠AXB = ∠XAB

AB= BX (sides opposite of equal angles)

Now,AB/AD= BX/BC

=AB/AD=BX/2BX

(since,X is mid point of BC)

=AB/AD=1/2

= AD = 2AB