4. Given an isosceles triangle $A B C$, where $\angle A=30^{\circ}, A B=A C$. Point $D$ is the midpoint of $B C$. On segment $A D$, point $P$ is chosen, and on side $A B$, point $Q$ is chosen such that $P B=P Q$. What is the measure of angle $P Q C ?$ (S. S. Korechkova)
Answer: $15^{\circ}$.

Solution. Since $D$ is the midpoint of the base of the isosceles triangle, $A D$ is the median, bisector, and altitude of the triangle. Draw the segment $P C$. Since $\triangle P D B = \triangle P D C$ (by two sides and the right angle between them), $P C = P B = P Q$, meaning that all three triangles $\triangle P B C$, $\triangle P B Q$, and $\triangle P Q C$ are isosceles.

Notice that $\angle A B D = \angle A C B = (180^{\circ} - 30^{\circ}) : 2 = 75^{\circ}$. Let $\angle P B D = \angle P C D = \alpha$, then $\angle P B Q = \angle P Q B = 75^{\circ} - \alpha$, and $\angle P Q C = \angle P C Q = \beta$. The sum of the angles in triangle $\triangle Q C B$ is $2 \alpha + 2 \cdot (75^{\circ} - \alpha) + 2 \beta = 180^{\circ}$, from which $\beta = 15^{\circ}$.

Criteria. Correct answer without justification - 0 points.

Proven that $P B = P C = P Q - 3$ points.

![](https://cdn.mathpix.com/cropped/2024_05_06_758bf2042e6fda9db2ecg-2.jpg?height=542&width=411&top_left_y=180&top_left_x=1508)