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Question
In the figure, given below, AM bisects angle A and DM bisects angle D of parallelogram ABCD. Prove that: ∠AMD = 90°.
Solution
From the given figure we conclude that
∠A + ∠D = 180° ...[Since consecutive angles are supplementary ]
`(∠A )/2 + (∠ D)/2 `= 90°
Again from the ΔADM
`(∠A) /2 + (∠ D)/2 ` + ∠M = 180°
⇒ 90° + ∠M = 180° ...`[since (∠A) /2 + (∠ D)/2 = 90° ]`
⇒ ∠M = 90°
Hence ∠AMD = 90°
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