Question

In a parallelogram ABCD the bisector of angle A meets DC in E and AB=2AD. Prove that BD bisects Angle B and angle AEB is a right angle

Answer 1

                  (i)  Let AD = xAB = 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,

Answer 2

Answer:

Step-by-step explanation:

    (i)  Let AD = xAB = 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,

HOPE IT IS USEFUL

THANK U