Question

In the adjoining figure ABCD is a parallelogram in which angel A=60°.If the bisectors of angel A and angel B meet DC at p , prove that angel APB=90°​

Answer 1

Answer:

In parallelogram opposite angles are equal

so, angle A=AngleC

let angle D=B be x

x+x+60°+60°=360°(angle sum property of parallelogram)

2x+120°=360°

2x=360°-120°

2x=240°

x=240/2

x=120°

AngleD=angleB=120°

angle A=60°

AP bisect angle A

angle PAB=30°( half of 60°)

BP bisect angle B

anglePBA=60°(half of 120°)

60°+30°+angle APB=180°( Angle sum property of triangle)

90°+angle APB=180°

angle APB=180°-90°

angleAPB=90°

Proved....

in the above where I have used sum of all angles in parallelogram is 360°

you can use sum of two adjacent angles in a parolllelogram is equal to 180°

you can use this

60°+x=180°

x=180°-60°

x=120°

here x =angleD

angleD=angleB( opposite angles are equal in parallelogram)