Question

2. AM bisects angle
A and DM bisects angle D of parallelogram
ABCD. Prove that : ZAMD = 90°​

Answer 1

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Answer 2

Given : M bisects angle A and DM bisects angle D of parallelogram

To prove : ∠AMD = 90°

Proof :

$\angle A+ \angle D = 180^\circ \\\\\frac{\angle A}{2}+ \frac{\angle D}{2}= 90^\circ \\\\\text{in triangle ADM }\\\\\frac{\angle A}{2}+ \frac{\angle D}{2}+\angle M = 180^\circ\\\\since , \\\\\frac{\angle A}{2}+ \frac{\angle D}{2}= 90^\circ\\\\90^\circ +\angle M = 180^\circ \\\\\angle M = 90^\circ \\\\\angle AMD = 90^\circ$

hence  proved

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