Question

9. 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​

Answer 1

Given :

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

To prove :

• (i) BP bisects angle B.
• (ii) Angle APB = 90°.

Solution :

i]

(1) In || gm ABCD AP is bisector of ∠A

∠1 = ∠2

(2) AB || DC and AP is the transversal,

∠2 = ∠3 (Alternate angles)

(3) ∠1 = ∠3 From (1) and (2)

(4) In ΔADP, we have

∠1= 2∠3 Proved in (3)

DP=AD

[Sides opposite to equal angles are equal]

AD = DP

(5) AB = 2AD........ (Given)

AB/2 = AD

AD = AB/2

(6) AB || DC

[Opposite sides of || gm are equal]

(7) AD = DC/2

(8) DP = DC/2 From (4) and (7)

P is the mid-point of DC.

DP = PC

(9) AD = BC

[Opposite sides of || gm are equal]

(10) DP = BC From (4) and (9)

(11) PC = BC From (8) and (10)

(12) In ΔBCP, we have

PC = BC Proved in (11)

∠6 = ∠5

[Angles opposite to equal sides are equal.]

(13) AB || DC

[ABCD is a || gm and BP is the transversal ]

∠4 = ∠5 (alternate ∠s)

(14) ∠4 = ∠6 From (12) and (13)

BP is bisector of ∠B.

ii]

(15) AD || BC

[Opposite sides of gm ABCD and AB

is the transversal]

∠A + ∠B = 180°

[Sum of co-interior angles = 180°]

=> ½(∠A+ ∠B) = ½ x 180

=> ∠APB = 90°