Question

in a parallelogram ABCD the bisectors of angle A and Angle B intersect each other at a point. Prove that angle APB is equal to 90 degree​

Answer 1

let P Q R and S be the points of intersection of the bisectors of angle A and Angle B Angle B and angle C angle C and angle D and Angle B and angle respectively of parallelogram ABCD

in a triangle ASD since DS bisects angle D and AD bisects angle A therefore angle d a s + + angle A D S is equal to one by two angle A + 1 by 2 angle d is equal to 1 by 2 angle A + angle d is equal to 1 by 2 is equal to 1 by 2 into 180 as angle A and Angle B are interior angles on the same side of the transversal is equal to 90 degree also angle d a s + angle A + angle d is equal to 180 angle sum property 90 + angle d is equal to 180 DS is equal to 90 angle BAC is equal to 90 being vertical vertically opposite angle d s a