3. Given an acute scalene triangle $A B C$. In it, the altitudes $B B_{1}$ and $C C_{1}$ are drawn, intersecting at point $H$. Circles $\omega_{1}$ and $\omega_{2}$ with centers $H$ and $C$ respectively touch the line $A B$. From point $A$ to $\omega_{1}$ and $\omega_{2}$, tangents other than $A B$ are drawn. Denote the points of tangency with these circles by $D$ and $E$ respectively. Find the angle $B_{1} D E$.
Answer: $180^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_c43b1f7fb831d10d42ecg-07.jpg?height=497&width=516&top_left_y=1599&top_left_x=790)

Solution. Since right triangles $A C C_{1}$ and $A C E$ are equal by leg and hypotenuse, line $C A$ is the bisector of isosceles triangle $C_{1} C E$. Therefore, it is the perpendicular bisector of segment $C_{1} E$, meaning points $C_{1}$ and $E$ are symmetric with respect to line $A C$. From the equality of triangles $A H D$ and $A H C_{1}$, it follows that $\angle A H D = \angle A H C_{1}$. The quadrilateral $A B_{1} H C_{1}$ has two opposite right angles. Therefore, it is cyclic, from which $\angle A B_{1} C_{1} = \angle A H C_{1}$. By the symmetry of points $C_{1}$ and $E$,

$$
\angle A B_{1} E = \angle A B_{1} C_{1} = \angle A H C_{1}.
$$

Moreover, $\angle A D H = 90^{\circ} = \angle A B_{1} H$. Therefore, quadrilateral $A D B_{1} H$ is also cyclic, and

$$
\angle A B_{1} D = \angle A H D = \angle A H C_{1}
$$

Thus, $\angle A B_{1} D = \angle A B_{1} E$, which means $\angle B_{1} D E = 180^{\circ}$.