3. A circle $\omega$ is circumscribed around triangle $ABC$. A line tangent to $\omega$ at point $C$ intersects the ray $BA$ at point $P$. On the ray $PC$ beyond point $C$, a point $Q$ is marked such that $PC = QC$. The segment $BQ$ intersects the circle $\omega$ again at point $K$. On the smaller arc $BK$ of the circle $\omega$, a point $L$ is marked such that $\angle LAK = \angle CQB$. Find the angle $\angle PCA$, given that $\angle ALQ = 60^{\circ}$.
Answer: $30^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_9304f7a3bc6f36cb1f9ag-37.jpg?height=449&width=462&top_left_y=398&top_left_x=931)

Solution. Inscribed angles $L A K$ and $L B K$ subtend the arc $L K$ and are therefore equal. Given that $\angle L B Q = \angle B Q C$, it follows that $B L \| P Q$. Then the line $\ell$, passing through point $C$ and perpendicular to the tangent $P Q$, contains the diameter of the circle $\omega$ and is perpendicular to its chord $B L$. Therefore, points $B$ and $L$ are symmetric with respect to $\ell$, meaning that $\ell$ is the axis of symmetry of the trapezoid $P Q L B$. From this, it follows that $B P = L Q$ and $B C = L C$. Hence, triangles $Q L C$ and $P B C$ are congruent by the three sides, so $\angle Q L C = \angle A B C$. Additionally, $\angle A L C = \angle A B C$ as inscribed angles subtending the same chord. Therefore, $L C$ is the angle bisector of $\angle A L Q$, and

$$
\angle A B C = \angle A L C = \frac{1}{2} \angle A L Q = 30^{\circ}.
$$

Note that the angle between the chord $A C$ and the tangent $C P$ is equal to the inscribed angle subtending $A C$. Therefore, $\angle P C A = \angle A B C = 30^{\circ}$.