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