Question

ABCDE is a regular pentagon.The bisector of <A meets CD at M .Find

<AMC​

Answer 1

Given :

1. ABCDE is a regular pentagon.
2. 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 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°

Hence Proved

Please

F.O.L.L.O.W

M.E.