Problem 5. In triangle $A B C$, the angle at vertex $B$ is $120^{\circ}$, point $M$ is the midpoint of side $A C$. Points $E$ and $F$ are chosen on sides $A B$ and $B C$ respectively such that $A E=E F=F C$. Find $\angle E M F$.
Answer: $90^{\circ}$.

Solution. Note that the sum of angles $A$ and $C$ is $60^{\circ}$. Let $A E=E F=F C=u$.

On the line $E M$, place a point $G$ such that point $M$ is the midpoint of segment $E G$. Triangles $A M E$ and $C M G$ are equal by two sides and the angle between them, so $\angle M C G=\angle M A E=\angle A$ and $C G=A E=u$.

Note that $\angle F C G=\angle F C M+\angle M C G=\angle C+\angle A=60^{\circ}$.

Since $F C=u=C G$, triangle $F C G$ is equilateral, and $F G=u$. Therefore, triangle $E F G$ is isosceles, $E F=F G=u$. Its median $F M$ is also an altitude, so $\angle E M F=90^{\circ}$.

## Criteria

One of the largest suitable criteria is used:

7 p. Any complete solution to the problem.

1 p. A point symmetric to $E$ or $G$ relative to $M$ is considered, but there is no further progress.

0 p. Only the correct answer is given.