4. In an acute-angled triangle $A B C$, the altitudes $A A_{1}, B B_{1}$, and $C C_{1}$ are dropped. A point $T$ is chosen on the plane such that the lines $T A$ and $T B$ are tangents to the circumcircle of triangle $A B C$, and point $O$ is the center of this circle. The perpendicular dropped from point $T$ to the line $A_{1} B_{1}$ intersects the line $C C_{1}$ at point $K$, and the line passing through point $C_{1}$ parallel to $O K$ intersects the segment $C O$ at point $L$. Find the angle $\angle C L A_{1}$.
Answer: $90^{\circ}$

![](https://cdn.mathpix.com/cropped/2024_05_06_e0197926f6acc0f95da5g-08.jpg?height=1217&width=966&top_left_y=545&top_left_x=545)

Solution. Let $\angle B A C=\alpha$ and $\angle A C B=\gamma$. Triangles $A O T$ and $B O T$ are congruent by three sides, and the inscribed angle $\angle A C B$ subtends the same arc as the central angle $\angle A O B$, so $\gamma=\angle A C B=\frac{1}{2} \angle A O B=\angle T O B$. Therefore, $\frac{O B}{O T}=\cos \gamma$. Similarly, $\angle B O C=2 \angle B A C$ and from the isosceles property of triangle $B O C$, it follows that $\angle B C O=90^{\circ}-\frac{1}{2} \angle B O C=90^{\circ}-\angle B A C$. Since $\angle A B_{1} B=90^{\circ}=\angle A A_{1} B$, quadrilateral $A B_{1} A_{1} B$ is cyclic. Therefore, $\angle B A C=\angle C A_{1} B_{1}$. Thus, $\angle B C O=90^{\circ}-\angle B A C=90^{\circ}-\angle C A_{1} B_{1}$ and lines $A_{1} B_{1}$ and $C O$ are perpendicular. Therefore, lines $C O$ and $T K$ are parallel, and quadrilateral $C O T K$ is a parallelogram, in particular, $C K=O T$. Then, by Thales' theorem, $\frac{C L}{C C_{1}}=\frac{C O}{C K}=\frac{O B}{O T}=\cos \gamma$. Therefore,

$$
C L=C C_{1} \cos \gamma=A C \sin \alpha \cos \gamma=C A_{1} \sin \alpha=C A_{1} \sin \angle C A_{1} B_{1} .
$$

Thus, point $L$ is the foot of the perpendicular dropped from $C$ to line $A_{1} B_{1}$, meaning $\angle C L A_{1}=90^{\circ}$.