Question

In a parallelogram ABCD, the bisectors of adjacent angles A and D intersect each other at
point P. Prove that APD  90o

Answer 1

Answer:

Answer is no more than a day after the fact

Step-by-step explanation:

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Answer 2

Given,

parallelogram ABCD , the

bisectors of consecutive angles /_A and

/_B intersect at p

The AD || BC and AB transversal then

.

. . /_ DAB + /_CAB = 180°

=> _1 /_DAB +/_CBA=180°

2

the bisectors of consecutive angles

/_A and /_B intersect at p

Then /_PAB = _1 /_DAB and /_PAB =_1

2 2

/_CAB

.

. . /_PBA + /_PBA = 90° in ∆ ABT

/_APB + /_PBA + /_PBA = 180°

=> /_APB + 90° = 180°

=> /_APB = 180 = 90 = 90°