Question

Show that the bisector of angles of a parallelogram form a rectangle

Answer 1

Angle A + Angle D = 180° As the adjacent angles of ||gm are supplementary.

Then
1/2 Angle A + 1/2 Angle D = 90°
Now
In Triangle ASD half of angle A + half of angle D + angle S = 180° ( Angle sum property)
Angle S = 90°. (1/2 Angle A + 1/2 Angle D = 90°)
Similarly we can prove
Angle Q = 90°
Now
PQ || SR. ( Both lines are perpendicular to AP And CR)
therefore angle B = angle C = 90°
hence PQRS is a rectangle
Proved..

Answer 2

let P,Q,Rand S be the points of intersecting of the bisector of angle A and angle B, angle B and angle C, angle C and angle D , and angle D and angle A respectively of a parallelogram ABCD

in triangle ASD

since DS bisects angle D and AS bisects angle A, therefore,

angle DAS +angle ADS=1/2 angle A+1/2 angle D

=1/2 {angleA+angleD}

= 1/2 x 180 ( angle A and angle D are interior angles on the same side of the transversal)

= 90

Also, angle DAS +   angle ADS +  angleDSA =180 ( angle sum property of a triangle)

Or,

90+  angle DSA = 180

Or,

angle DSA= 90

So,

angle psr =90  (being vertically opposite to  angle DSA)