Question

c) In the parallelogram ABCD, the bisector of angles A and D intersect each other at P and angle DAB = 50°.
Prove that angle APD = 90°.

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Answer 1

Answer:

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Step-by-step explanation:

In parallelogram ABCD , angle A = 80° . Bisectors of angle A and B are AM and

BM meet at point M.

In triangle MAB , angle MAB = 1/2 of angle A = 1/2× 80° =40°

And angle MBA= 1/2.of angle B. = 1/2×(180–80)° = 50°.

Angle MAB+angle MBA +angle AMB= 180°

or. 40°+50°+ angle AMB =180°

or. Angle AMB=180°-90° = 90°. Answer.