8,9 |
| :---: | :---: | :---: |
|  | Inscribed and Circumscribed Circles |  |
|  | [ Properties and characteristics of isosceles triangles.] $]$ |  |

In triangle $A B C$, $\angle B=36^{\circ}, \angle C=42^{\circ}$. On side $B C$, a point $M$ is taken such that $B M=R$, where $R$ is the radius of the circumscribed circle of triangle $A B C$.

Find the angle $MAC$.
Prove that point $M$ lies on the radius connecting the center of the circumscribed circle with vertex $A$.

## Solution

Let $O$ be the center of the circumscribed circle. $\angle A O B = 2 \angle C = 84^{\circ}, \angle B O C = 2 \angle B + 2 \angle C = 156^{\circ}, \angle O B C = 90^{\circ} - 1 / 2$ $\angle B O C = 12^{\circ}$,

$\angle B O M = 90^{\circ} - 1 / 2 \angle O B M = 84^{\circ} = \angle B O A$.

Therefore, point $M$ lies on the segment $O A$. Thus, $\angle M A C = \angle B A C - \angle B A O = 54^{\circ}$.

## Answer