Question

The bisectors of two adjacent sides of a parallelogram ABCD meet at a point P inside the parallelogram. The angle made by these bisectors at point P is ____________.

Answer 1

Given:
ABCD is a parallelogram
Bisectors of ∠A and ∠B intersect each other at P


∠A + ∠B = 180° (interior angles)
⇒ $\frac{1}{2}$∠A + $\frac{1}{2}$∠B = $\frac{1}{2}$(180°)
⇒ ∠PAB + ∠PBA = 90° ....(1)

Now, in ∆APB
∠PAB + ∠PBA + ∠APB = 180° (angle sum property)
⇒ 90° + ∠APB = 180° (From (1))
⇒ ∠APB = 180° − 90°
⇒ ∠APB = 90°


Hence, the angle made by these bisectors at point P is 90°.