Given a triangle $A B C$, in which $A B=A C$ and $\angle A=110^{\circ}$. Inside the triangle, a point $M$ is taken, such that $\angle M B C=$ $30^{\circ}$, and $\angle M C B=25^{\circ}$. 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 $A$. Since $\angle B N C=120^{\circ}$, then $\angle A N C=\angle A N B$ $=120^{\circ}$, and since $\angle C=\angle B=35^{\circ}$,

$\angle A C N=\angle A B N=5^{\circ}$, then $\angle N C M=35^{\circ}-25^{\circ}-5^{\circ}=5^{\circ}$.

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

Thus, $A C=C M$,

$\angle A M C=\angle M A C=1 / 2\left(180^{\circ}-\angle A C M\right)=1 / 2\left(180^{\circ}-10^{\circ}\right)=85^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_e3b1ddf947c7dd2094eeg-04.jpg?height=363&width=1172&top_left_y=1711&top_left_x=449)

Answer

$85^{\circ}$.

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