Question

In parallelogram ABCD, the bisector of angle
A meets DC at P and AB = 2AD.
Prove that:
(i) BP bisects angle B. (ii) Angle APB = 90°​

Answer 1

Answer:

ANSWER

(i) Let AD = AB = 2AD = 2x

Also AP is the bisector ∠A∴∠1 = ∠2

Now, ∠2 = ∠5 (alternate angles)

∴∠1 = ∠5Now AD = DP = x [∵ Sides opposite to equal angles are also equal]

∵ AB = CD (opposite sides of parallelogram are equal)

∴ CD = 2x⇒ DP + PC = 2x⇒ x + PC = 2x⇒ PC = x

Also, BC = x In ΔBPC,∠6 = ∠4 (Angles opposite to equal sides are equal)

Also, ∠6 = ∠3 (alternate angles)

∵ ∠6 = ∠4 and ∠6 = ∠3⇒∠3 = ∠4

Hence, BP bisects ∠B.

(ii) To prove ∠APB = 90°∵ Opposite angles are supplementary..

Angle sum property,