Answer 1

Given: ABCDE is a regular pentagon.

The bisector ∠A of the pentagon meets the side CD at point M.

To prove : ∠AMC = 90°

Proof: We know that the measure of each interior angle of a regular pentagon is 108°.

∠BAM =  x 108° = 54°

Since, we know that the sum of a quadrilateral is 360°

In quadrilateral ABCM, we have

∠BAM + ∠ABC + ∠BCM + ∠AMC = 360°

54° + 108° + 108° + ∠AMC = 360°

∠AMC = 360° – 270°

∠AMC = 90°