Question

ABCDE is a regular pentagon. The bisector of angle A meets the side CD at M. Find ∠AMC.

Answer 1

Given, A pentagoan ABCDE.

The line segment AM is the bisector of the ∠A.

Now, Since the measure of each interior angle of a regular pentagon is 108°.

∴ ∠BAM = `1/2 xx 108^circ` = 54°

By the angle sum property of a quadrilateral, we have (in quadrilateral ABCM)

∠BAM + ∠ABC + ∠BCM + ∠AMC = 360°

⇒ 54° + 108° + 108° + ∠AMC = 360°

⇒ ∠AMC = 360° – 270°

⇒ ∠AMC = 90°