Question

ABCDE is a regular pentagon. The bisector of angle A of the pentagon meets the side CD in point M. Show that $\angle AMC={90}^o$.

Answer 1

Given : ABCDE is a regular pentagon. The bisector $\angle A$ of the pentagon meets the side CD at point M.

To prove : $\angle AMC={90}^o$

Proof : We know that, the measure of each interior angle of a regular pentagon is ${108}^o$.

$\therefore \angle BAM=\frac{1}{2}\times {108}^o={54}^o$

Since, we know that the sum of a quadrilateral is ${360}^o$

$\therefore \text{In quadrilateral ABCM, we have}\angle BAM+\angle ABC+\angle BCM+\angle AMC={360}^o{54}^o+{108}^o+{108}^o+\angle AMC={360}^o\angle AMC={360}^o-{270}^o\angle AMC={90}^o$.