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.

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