Abcde is a Regular Pentagon. the Bisector of Angle a of the Pentagon Meets the Side Cd in Point M. Show that ∠Amc = 90°. - Mathematics

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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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Chapter 16: Understanding Shapes - Exercise 16 (C) [Page 188]

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Selina Concise Mathematics Class 8 ICSE
Chapter 16 Understanding Shapes
Exercise 16 (C) | Q 14 | Page 188

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