Question

ABCD is a parallelogram in which the bisectors of angle A and angle B intersect at a Point P. prove that angle APB=90degree.

Answer 1

Since ABCD is a Parallelogram. Therefore,
AD || BC
AB is a transversal . Therefore ,

A + B = 180°. [ Consecutive interior angles]

Multiply both sides by 1/2 ,

1/2 A + 1/2 B = 1/2 (180°)
1/2 A + 1/2 B = 90° __【1】

Since, AP and PB are angle bisectors of A and B . Therefore,

Angle 1 = 1/2 A
Angle 2 = 1/2 B

Substitute the values in【1】,

Angle 1 + Angle 2 = 90°____【2】

Now, in ∆ APB,
1 + APB + 2 = 180°
90° + APB = 180°. [From 【2】]
APB = 90°
- HENCE PROVED

Hope this helps :)

Answer 2

Answer :
In a parallelogram ABCD, ∠1 = ∠2, and ∠3 = ∠4
To Prove: ∠APB = 90°
Proof: Since AD ïï BC with transversal AB,  
... ∠A + ∠B = 180°(Consecutive Interior angles) ⇒∠A + ∠B = 90° ⇒ ∠2 + ∠3 = 90°  ...... (i)
Now in ΔABP, 
we have 
∠2 + ∠3 + ∠APB = 180° ...... (ii) ⇒ 90° + ∠APB = 180°
... ∠APB = 180° - 90° = 90°