3. Given a triangle $ABC$ with the largest side $BC$. The bisector of its angle $C$ intersects the altitudes $AA_{1}$ and $BB_{1}$ at points $P$ and $Q$ respectively, and the circumcircle of $ABC$ at point $L$. Find $\angle ACB$, if it is known that $AP=LQ$.


Answer: $60^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_66368caef28b3de8e403g-23.jpg?height=503&width=528&top_left_y=979&top_left_x=821)

Solution. Let $\alpha=\angle B C L, \beta=\angle A L C, \gamma=\angle B L C$ (see the figure). We will prove the equality of triangles $A L P$ and $B L Q$. Note that $A P=L Q$ by the condition and $A L=L B$ as chords corresponding to equal angles. Moreover,

$$
\angle A P L=\angle A_{1} P C=90^{\circ}-\alpha=\angle B_{1} Q C=\angle B Q L
$$

Then, by the Law of Sines,

$$
\frac{A P}{\sin \angle A L P}=\frac{A L}{\sin \angle A P L}=\frac{L B}{\sin \angle B Q L}=\frac{L Q}{\sin \angle L B Q}, \quad \text { hence } \quad \sin \angle A L P=\sin \angle L B Q
$$

But $\angle A L P=\angle A B C<90^{\circ}$ and

$$
\angle L B Q=180^{\circ}-\angle B L C-\angle B Q L=90^{\circ}-\angle B A C+\alpha<90^{\circ}-\angle B A C+\angle A C B \leqslant 90^{\circ}
$$

(the last inequality is true because $B C \geqslant A B$). Therefore, the angles $A L P$ and $L B Q$ are acute, and they are equal due to the equality of their sines. Thus, triangles $A L P$ and $B L Q$ are equal by a side and two angles. Therefore, $\angle L A P=\angle Q L B=\gamma$. Now from triangle $A L P$

$$
\beta+\gamma+90^{\circ}-\alpha=180^{\circ} \Longrightarrow \beta+\gamma=90^{\circ}+\alpha
$$

and from quadrilateral $A L B C$

$$
\beta+\gamma=180^{\circ}-2 \alpha \Longrightarrow 180^{\circ}-2 \alpha=90^{\circ}+\alpha \Longrightarrow \alpha=30^{\circ} \Longrightarrow \angle A C B=60^{\circ}
$$