7. (40 points) $Q L$ is the bisector of triangle $P Q R$, and $M$ is the center of the circumcircle of triangle $P Q L$. It turned out that points $M$ and $L$ are symmetric with respect to $P Q$. Find the angles of triangle $P Q L$.

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Answer: $\angle P L Q=120^{\circ}, \angle L P Q=\angle P Q L=30^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_9d0c3613b9652189bb5dg-6.jpg?height=388&width=548&top_left_y=991&top_left_x=754)

Solution: Let segments $M L$ and $P Q$ intersect at point $H$. According to the problem, $M L \perp P Q$ and $M H=H L$. Therefore, right triangles $M H P$ and $H P L$ are equal by two legs; similarly, right triangles $M H Q$ and $H Q L$ are equal. Since $M$ is the center of the circumscribed circle around triangle $P Q L$, then $M P=M Q=M L$. Thus, $P L=M P=M Q=Q L=M L$, and triangles $M P L$ and $M Q L$ are equilateral. From this, we get that $\angle P L Q=120^{\circ}$. It is also easy to notice that triangle $P L Q$ is isosceles, so $\angle Q P L=\angle P Q L=\left(180^{\circ}-\angle P L Q\right) / 2=30^{\circ}$.