Question

In parallelogram ABCD, the bisector of angle A meets DC at P and AB = 2 AD.
Prove that:
(i) BP bisects angle B.
(ii) Angle APB = 90^o.

Answer 1

(i) Let AD = x
AB = 2AD = 2x
Also AP is the bisector ∠A
∠1 = ∠2
Now,
∠2 = ∠5                   ...[ alternate angles ]
Therefore ∠1 = ∠5
Now
AP = DP = x ...[ sides opposite to equal angles are also equal ]
Therefore
AB = CD ...[ opposite sides of  parallelogram are equal ]
CD = 2x
⇒ DP + PC = 2x
⇒ x + PC = 2x
⇒ PC = x
Also, BC = x
ΔBPC
⇒ ∠6 = ∠4 ...[ angles opposite to equal sides are equal ]
⇒ In ∠6 = ∠3
Therefore ∠3 =∠ 4
Hence BP bisect ∠B

(ii)
Opposite angles are supplementary
Therefore

∠1 + ∠2 + ∠3 + ∠4 = 180°
⇒ 2 ∠2 + 2 ∠3 =180°     .....[ ∠1 = ∠2 , ∠3 = ∠4 ]
⇒ ∠2 + ∠3 = 90°
ΔAPB
∠2 + ∠3 ∠APB = 180°
⇒ ∠APB = 180° - 90° ...[ by angle sum property ] 
⇒ ∠APB = 90° 
Hence proved.