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 °.
Δ
(i) Let AD = x
AB = 2AD = 2x
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 = 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.
In ΔAPB;
By Angle sum property,
Hence proved.