Question

let ABCD be a parallelogram on suppose the bisector of angle A and Angle B meet at P prove that angle APB = 90 degree

Answer 1

I hope you will understand

Answer 2

We know that,
angle A & B= 90°+90° = 180°(sum of angle at the one side of parallelogram is 180°

In ∆ABP,
angle A + angle B + angle P = 180° (Sum of the angles of a triangle is 180°)
= 90° + angle P = 180°
=angle P = 180°–90°
=angle P = 90°
Therefore angle APB = 90°

I hope it's clear.
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