Question

In a parallelogram ABCD, the bisectors of adjacent angles A and D intersect each other at point P. Prove that ∠APD = 90° what will be the diagram for this question ??

Answer 1

Given : In a parallelogram ABCD, the bisectors of adjacent angles A and D intersect each other at point P.

To Find : Prove that ∠APD = 90°  

Solution:

Opposite angles of a parallelogram are equal

Sum of adjacent angles of a parallelogram = 180°

=> ∠A + ∠D = 180°

=> (1/2)(∠A +∠D) = 90°

AP is angle bisector of ∠A

∠DAP =∠A/2

DP is angle bisector of ∠D

∠ADP =∠D/2

∠DAP + ∠ADP + ∠APD = 180°  ( sum of angles of  a triangle )

=> ∠A/2 + ∠D/2 + ∠APD = 180°

=> (1/2)(∠A +∠D)   + ∠APD = 180°

=> 90°  + ∠APD = 180°

=> ∠APD = 90°

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