Question

In the adjoining figure, ABCD is a parallelogram in which the bisectors of ​∠A and ​∠B intersect at a point P. Prove that ​∠APB = 90°.

Answer 1

ABCD is a parallelogram and ∠A and ∠B are adjacent angles.
∴ ​∠ A + ∠B = 180^o.
⇒ ​$\frac{\angle A}{2}+\frac{\angle B}{2}=\frac{180}{2}={90}^o....................\left(i\right)$
In ∆​ APB, we have:
∠​PAB = ∠A​ /2
∠​PBA = ∠B​ /2​
∴ ∠​APB = 180 − (∠​PAB​ + ∠​PBA​) [ Angle sum property of triangle]
⇒ ​∠APB = 180 − $\left(\frac{\angle A}{2}+\frac{\angle B}{2}\right)$
⇒ ∠APB = 180 − 90 = 90^o
Hence, proved.