7. 

In triangle $A B C$, the bisector $A L$ ( $L \in B C$ ) is drawn, and $M$ and $N$ are points on the other two bisectors (or their extensions) such that $M A=M L$ and $N A=N L, \angle B A C=50^{\circ}$.

Find the measure of $\angle M A N$ in degrees.
# Solution

We will use the auxiliary statements.

If the bisector $B K$ in triangle $A B C$ intersects the circumscribed circle at point $W$, then:

1) $A W=C W$ (since $\angle C A W=\angle C B W=\angle A B W=\angle A C W$, that is, triangle $A W C$ is isosceles and $A W=C W$).

![](https://cdn.mathpix.com/cropped/2024_05_06_17255c54865c30865cf7g-4.jpg?height=469&width=422&top_left_y=902&top_left_x=977)

2) Point $W$ lies on the perpendicular bisector of segment $A C$.
3) The bisector of the angle of a non-isosceles triangle and the perpendicular bisector of the opposite side intersect on the circumscribed circle.

Next, consider triangle $A B L (A B \neq B L)$, $M N$ is the perpendicular bisector of $A L$ and $B N$ is the bisector of angle $B$. Therefore, points $A, B, L, N$ lie on the same circle.

![](https://cdn.mathpix.com/cropped/2024_05_06_17255c54865c30865cf7g-4.jpg?height=680&width=625&top_left_y=1810&top_left_x=858)

Similarly, points $A, C, L, M$ lie on the same circle.

Then:

$$
\begin{gathered}
\angle M A N=\angle L A M+\angle L A N=\angle L C M+\angle N B L= \\
=\frac{\angle C+\angle B}{2}=90^{\circ}-\frac{1}{2} \angle A=90^{\circ}-\frac{1}{2} \cdot 50^{\circ}=65^{\circ}
\end{gathered}
$$

Answer. 65.