$\left.\begin{array}{l}{[\text { Auxiliary equal triangles }} \\ \text { [ Congruent triangles. Criteria for congruence }]\end{array}\right]$

Given triangle $A B C$, where $A B=A C$ and $\angle A=80^{\circ}$. Inside triangle $A B C$, a point $M$ is taken such that $\angle M B C=30^{\circ}$, and $\angle M C B=10^{\circ}$. Find $\angle A M C$.
Let $N$ be the point of intersection of the line $B M$ with the bisector of angle $B A C$. Prove that triangles $A C N$ and $M C N$ are equal.

## Solution

Let $N$ be the point of intersection of the line $B M$ with the bisector of angle $B A C$. Since $\angle B N C=120^{\circ}$, then $\angle A N C=$ $\angle A N B=120^{\circ}$, and since $\angle C=\angle B=50^{\circ}$, $\angle A C N=\angle A B N=20^{\circ}$, then $\angle N C M=50^{\circ}-20^{\circ}-10^{\circ}=20^{\circ}$.

Therefore, triangles $A C N$ and $M C N$ are equal by side ( $C N$ is common) and two adjacent angles.

Thus, $A C=C M$,

$\angle A M C=\angle M A C=1 / 2\left(180^{\circ}-\angle A C M\right)=1 / 2\left(180^{\circ}-40^{\circ}\right)=70^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_f7f9d90261bfc416bd71g-05.jpg?height=397&width=728&top_left_y=2277&top_left_x=665)

## Answer

$70^{\circ}$.

## Problem