7. (30 points) Point $M$ is the midpoint of the hypotenuse $A C$ of the right triangle $A B C$. Points $P$ and $Q$ on lines $A B$ and $B C$ respectively are such that $A P = P M$ and $C Q = Q M$. Find the measure of angle $\angle P Q M$, if $\angle B A C = 17^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_f38d24aa746a26458142g-4.jpg?height=683&width=714&top_left_y=518&top_left_x=657)
Answer: $17^{\circ}$.

Solution. Since in the right triangle $ABC$ the angle $BAC=17^{\circ}$, the angle $BCA=73^{\circ}$. Note that triangle $QMC$ is isosceles with base $MC$, as by condition $QM=QC$. Similarly, triangle $PMA$ is isosceles with base $MA$. Therefore, the angles at the bases of these triangles are equal. From this, we get that $\angle QMP=180^{\circ}-\angle PMA-\angle QMC=180^{\circ}-17^{\circ}-73^{\circ}=90^{\circ}$.

Let points $O$ and $R$ be the feet of the perpendiculars dropped from points $Q$ and $P$ to the line $AC$, respectively. It is easy to notice that triangles $QOM$ and $MRP$ are similar by two angles. Since $CM=MA$, from the isosceles nature of triangles $QCM$ and $PMA$, it follows that

$$
MO=\frac{1}{2} CM=\frac{1}{2} MA=MR
$$

Let $x=MO$. From the right triangles $QOM$ and $PMR$, we get that $MQ=\frac{x}{\cos 73^{\circ}}$ and $MP=\frac{x}{\cos 17^{\circ}}$ respectively. Thus, for triangle $MQP$ we have

$$
\operatorname{tg} \angle PQM=\frac{MP}{MQ}=\frac{x}{\cos 17^{\circ}} \cdot \frac{\cos 73^{\circ}}{x}=\frac{\cos \left(\pi / 2-17^{\circ}\right)}{\cos 17^{\circ}}=\operatorname{tg} 17^{\circ}
$$

Since in the right triangle $MQP$ the angle $PQM$ is acute, from the last equality it follows that $\angle PQM=17^{\circ}$.