Question

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

Answer 1

Answer:

Step-by-step explanation:

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,