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Sum

ABCDE is a regular pentagon. The bisector of angle A of the pentagon meets the side CD in point M. Show that ∠AMC = 90°.

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#### Solution

**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°

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