Question

Question 159
ABCDE is a regular pentagon. The bisector of angle A meets the sides CD at M. Find $\angle AMC$

Answer 1

Given, a pentagon ABCDE. The line segment AM is the bisector of the $\angle A$.
Now, since the measure of each interior angle of a regular pentagon is ${108}^{\circ }$.
$\therefore \angle BAM=\frac{1}{2}\times {108}^{\circ }={54}^{\circ }$
By the angle sum property of a quadrilateral, we have (in quadrilateral ABCM)
$\angle BAM+\angle ABC+\angle BCM+\angle AMC={360}^{\circ }\Rightarrow {54}^{\circ }+{108}^{\circ }+{108}^{\circ }+\angle AMC={360}^{\circ }\Rightarrow \angle AMC={360}^{\circ }-{270}^{\circ }\Rightarrow \angle AMC={90}^{\circ }$.