Question

In the figure, given below, AM bisects angle A and DM bisects angle D of parallelogram ABCD. Prove that: ∠AMD = 90°^.

Answer 1

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°