Question

4. ABCDE is a regular pentagon. The bisector of
angle A of the pentagon meets the side CD in
point M. Show that /_AMC = 90°.​

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 = 1/2 X 108o = 54o 

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°