(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,