The points $M$ and $N$ are chosen on the angle bisector $A L$ of a triangle $A B C$ such that $\angle A B M=\angle A C N=23^{\circ} . X$ is a point inside the triangle such that $B X=C X$ and $\angle B X C=2 \angle B M L$. Find $\angle M X N$.
Answer: $\angle M X N=2 \angle A B M=46^{\circ}$.

Let $\angle B A C=2 \alpha$. The triangles $A B M$ and $A C N$ are similar, therefore $\angle C N L=\angle B M L=\alpha+23^{\circ}$. Let $K$ be the midpoint of the arc $B C$ of the circumcircle of the triangle $A B C$. Then $K$ belongs to the line $A L$ and $\angle K B C=\alpha$. Both $X$ and $K$ belong to the perpendicular bisector of the segment $B C$, hence $\angle B X K=\frac{1}{2} \angle B X C=\angle B M L$, so the quadrilateral $B M X K$ is inscribed. Then

$$
\angle X M N=\angle X B K=\angle X B C+\angle K B C=\left(90^{\circ}-\angle B M L\right)+\alpha=90^{\circ}-(\angle B M L-\alpha)=67^{\circ} .
$$

Analogously we have $\angle C X K=\frac{1}{2} \angle B X C=\angle C N L$, therefore the quadrilateral $C X N K$ is inscribed also and $\angle X N M=\angle X C K=67^{\circ}$. Thus, the triangle $M X N$ is equilateral and
$\angle M X N=180^{\circ}-2 \cdot 67^{\circ}=46^{\circ}$.

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