7.3. On side $B C$ of triangle $A B C$, a point $M$ is marked such that $A B=B M$ and $A M=M C$. It is known that angle $B$ is five times angle $C$. Find the angles of the triangle.
Answer. $\angle A=60^{\circ}, \angle B=100^{\circ}, \angle C=20^{\circ}$. Solution. Triangles $A B M$ and $AMC$ are isosceles, so the angles at their bases are equal. Let these angles be $x$ and $y$ respectively. Then, by the property of the exterior angle $A M B$ for triangle $AMC$, we have $x=2 y$. Therefore, the sum of angles $A$ and $C$ is $x+y+y=4 y=180^{\circ}-\angle B$. According to the condition $\angle B=5 y$ and thus $9 y=180^{\circ}$, which means $y=20^{\circ}$. Then $\angle A=3 y=60^{\circ}, \angle B=5 y=100^{\circ}$.