Question

In parallelogram ABCD, the angle bisector of ∠A bisects BC. Will angle bisector of B also bisect AD? Give reason.

Answer 1

Given, ABCD is a parallelogram, bisector of ∠A, bisects BC at F, i.e. ∠1 = ∠2, CF = FB.

Draw FE || BA.

ABFE is a parallelogram by construction   ...[∵ FE || BA]

⇒ ∠1 = ∠6   ...[Alternate angle]

But ∠1 = ∠2   ...[Given]

∴ ∠2 = ∠6

AB = FB   [Opposite sides to equal angles are equal] ...(i)

∴ ABFE is a rhombus.

Now, In ΔABO and ΔBOF,

AB = BF  ...[From equation (i)]

BO = BO   ...[Common]

AO = FO  ...[Diagonals of rhombus bisect each other]

∴ ΔABO ≅ ΔBOF   ...[By SSS]

∠3 = ∠4   ...[By CPCT]

Now, BF = `1/2` BC  ...[Given]

⇒ BF = `1/2` AD   ...[BC = AD]

⇒ AE = `1/2` AD   ...[BF = AE]

∴  E is the midpoint of AD.