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