In the following figure, ABCD is a parallelogram. Prove that :

(i) AP bisects angle A

(ii) BP bisects angle B

(iii) ∠DAP + ∠CBP = ∠APB

(i) In △ ADP,

⇒ AD = DP (Given)

∴ ∠APD = ∠PAD (Angles opposite to equal sides are equal) …….(1)

In parallelogram ABCD,

AB || DC and AP is the transversal.

∴ ∠APD = ∠PAB (Alternate angles are equal) ……..(2)

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

⇒ ∠PAD = ∠PAB.

∴ AP bisects angle A.

Hence, proved that AP bisects angle A.

(ii) In || gm ABCD,

⇒ BC = AD (Opposite sides of || gm are equal)

∴ BC = PC

∴ ∠BPC = ∠PBC (Angles opposite to equal sides are equal) …….(3)

In parallelogram ABCD,

AB || DC and BP is the transversal.

∴ ∠BPC = ∠PBA (Alternate angles are equal) ……..(4)

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

⇒ ∠PBC = ∠PBA.

∴ BP bisects angle B.

Hence, proved that BP bisects angle B.

(iii) We know that,

Consecutive angles of a || gm are supplementary.

∴ ∠A + ∠B = 180°,

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

⇒ $\dfrac{∠A + ∠B}{2}$ = 90°

⇒ $\dfrac{∠A}{2} + \dfrac{∠B}{2}$ = 90°

⇒ ∠PAB + ∠PBA = 90° ………(1)

In △ APB,

⇒ ∠PAB + ∠PBA + ∠APB = 180°

⇒ 90° + ∠APB = 180°

⇒ ∠APB = 180° - 90°

⇒ ∠APB = 90° ……..(2)

From figure,

⇒ ∠PAB = ∠DAP (As, AP bisects ∠A)

⇒ ∠PBA = ∠CBP (As, BP bisects ∠B)

Substituting value of ∠PAB and ∠PBA in equation (1), we get :

⇒ ∠DAP + ∠CBP = 90° ………(3)

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

⇒ ∠DAP + ∠CBP = ∠APB.

Hence, proved that ∠DAP + ∠CBP = ∠APB.