Q1 of 40 Page 8

In a parallelogram ABCD, bisectors of consecutive angles A and B intersect at P. Prove that ∠APB = 90°.

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°

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