In a parallelogram ABCD, the bisector of angle A meets DC in P and AB = 2AD.

Prove that

1)BP bisects angle B.

2)Angle APB =90 °.

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(i) Let AD = *x*

AB = 2AD = 2*x*

Also AP is the bisector ∠A

∴∠1 = ∠2

Now, ∠2 = ∠5 (alternate angles)

∴∠1 = ∠5

Now AD = DP = *x* [∵ Sides opposite to equal angles are also equal]

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

∴ CD = 2*x*

⇒ DP + PC = 2*x*

⇒ *x* + PC = 2*x*

⇒ 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.

In ΔAPB;

By Angle sum property,

Hence proved.

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