Question

Let ABCD be a parallelogram and suppose the bisectors of ∠A and ∠B meet at P. Prove that ∠APB =90°.

Answer 1

Given :

ABCD is a parallelogram .

<A and <B bisectors meet at P.

To prove : <APO = 90°

Proof : In ABCD parallelogram ,

<A + <B = 180°

[ Sum of adjacent angles are

supplementary ]

=> <A/2 + <B/2 = 90°

=> <PAB + <ABP = 90° ----( 1 )

[ Angle bisectors of <A and <B ]

Now ,

In ∆APB,

<APB + <PAB + <ABP = 180°

[ Angle sum property ]

=> <APB + 90° = 180° [ from ( 1 ) ]

=> <APB = 180° - 90°

=> <APB = 90°

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