Question

Please help humble request ​

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°

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