In parallelogram ABCD, the bisectors of angle A meets DC at P and AB = 2AD.

Prove that :

(i) BP bisects angle B.

(ii) Angle APB = 90°.

(i) Let AD = x.

Given,

AB = 2AD = 2x.

Given,

AP is the bisector of angle A,

∴ ∠1 = ∠2 ………….(1)

From figure,

⇒ ∠2 = ∠5 [Alternate angles are equal] ……(2)

From equation (1) and (2), we get :

⇒ ∠1 = ∠5

In △ ADP,

⇒ ∠1 = ∠5

⇒ DP = AD = x (Sides opposite to equal angles are equal)

From figure,

⇒ AB = CD (Opposite sides of parallelogram are equal)

⇒ CD = 2x

⇒ DP + PC = 2x

⇒ x + PC = 2x

⇒ PC = 2x - x = x.

Also,

⇒ BC = AD = x (Opposite sides of parallelogram are equal)

In △ BCP,

⇒ BC = PC (Both equal to x)

⇒ ∠6 = ∠4 (Angles opposite to equal sides are equal) ……(3)

⇒ ∠6 = ∠3 (Alternate angles are equal) ……….(4)

From equation (3) and (4), we get :

⇒ ∠3 = ∠4.

∴ BP is the bisector of angle B.

Hence, proved that BP bisects angle B.

(ii) We know that,

Consecutive angles of parallelogram are supplementary.

Considering ∠A + ∠B = 180°,

⇒ $\dfrac{∠A + ∠B}{2} = \dfrac{180°}{2}$

⇒ $\dfrac{∠A + ∠B}{2}$ = 90°

⇒ $\dfrac{∠A}{2} + \dfrac{∠B}{2}$ = 90° …………(1)

In △ PAB,

By angle sum property of triangle,

⇒ ∠PAB + ∠ABP + ∠APB = 180°

⇒ $\dfrac{∠A}{2} + \dfrac{∠B}{2}$ + ∠APB = 180°

⇒ 90° + ∠APB = 180° [From equation (1)]

⇒ ∠APB = 180° - 90° = 90°.

Hence, proved that ∠APB = 90°.