3. In triangle $ABC$, angle $A$ is equal to $\alpha$, and $BC$ is the shortest side. On side $AB$, point $P$ is marked, and on side $AC$, point $Q$ is marked such that $PB = BC = CQ$. Segments $BQ$ and $CP$ intersect at point $M$. Find the measure of angle $BMC$.
Solution.

Triangles $P B C$ and $Q B C$ are isosceles, with corresponding sides marked on the diagram. Let the angles be: $\angle Q B C=\angle B Q C=x, \angle P B C=\angle B P C=$ $y, \angle A B Q=z, \angle A C P=v$. By the property of the exterior angle, applied to triangles $A B Q$ and $A P C$,

$$
\left\{\begin{array}{l}
\alpha+z=x \\
\alpha+v=y
\end{array}\right.
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_7735b71ca397ac3158a0g-2.jpg?height=400&width=809&top_left_y=128&top_left_x=652)

Write the sum of the angles of triangle $A B C$

$$
\alpha+z+x+y+v=180^{\circ}
$$

From the system, express $z+v$.

$$
z+v=x+y-2 \alpha
$$

Thus, $\alpha+2 x+2 y-2 \alpha=180^{\circ}, \quad x+y=90^{\circ}+\frac{1}{2} \alpha$.

Find the required angle

$$
\angle B M C=180^{\circ}-(x+y)=90^{\circ}-\frac{1}{2} \alpha
$$

Answer. $90^{\circ}-\frac{1}{2} \alpha$.