3. Given an acute-angled triangle $A B C$. A circle with diameter $B C$ intersects sides $A B$ and $A C$ at points $D$ and $E$ respectively. Tangents drawn to the circle at points $D$ and $E$ intersect at point $K$. Find the angle between the lines $A K$ and $B C$.
Answer: $90^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_627b343ad954e1e3a31bg-19.jpg?height=483&width=651&top_left_y=267&top_left_x=720)

Solution 1. Let $\alpha=\angle D B E, \beta=\angle B C D, \gamma=\angle C B E, O$ - the intersection point of segments $C D$ and $B E$. The angle between the tangent $D K$ and the chord $D E$ is equal to the inscribed angle that subtends

$$
\angle D E K=\angle E D K=\alpha
$$

Moreover, $\angle C D E=\gamma$. Due to the cyclic nature of quadrilateral $B D E C$

$$
\angle A E K=\angle A E D-\alpha=\angle C B D-\alpha=\gamma \quad \text { and, similarly } \quad \angle A D K=\beta .
$$

Since $\alpha+\beta+\gamma=\angle A D C=90^{\circ}$, we get

$$
\angle D A E=180^{\circ}-2 \alpha-\beta-\gamma=90^{\circ}-\alpha=\frac{1}{2}\left(180^{\circ}-2 \alpha\right)=\frac{1}{2} \angle D K E
$$

Note that a circle $\omega$ can be circumscribed around quadrilateral $A D O E$. Then the angle $D A E$ is inscribed in $\omega$, and point $K$ lies on the perpendicular bisector of chord $B E$, i.e., on the diameter of $\omega$. Therefore, $K$ is the center of $\omega$, from which

$$
\angle D A K=\angle A D K=\beta=90^{\circ}-\alpha-\gamma=90^{\circ}-\angle C B D
$$

Thus, $A K \perp B C$.

Solution 2. Let

$$
\alpha=\angle D B E, \quad \beta=\angle B C D, \gamma=\angle C B E, \varphi=\angle D A K, \quad \psi=\angle E A K
$$

The angle between the tangent $D K$ and the chord $D E$ is equal to the inscribed angle that subtends $D E$, hence

$$
\angle D E K=\angle E D K=\alpha
$$

Moreover, $\angle C D E=\gamma$. Due to the cyclic nature of quadrilateral $B D E C$

$$
\angle A E K=\angle A E D-\alpha=\angle C B D-\alpha=\gamma \quad \text { and, similarly } \quad \angle A D K=\beta .
$$

Since $\alpha+\beta+\gamma=\angle A D C=90^{\circ}$, we get

$$
\gamma=90^{\circ}-\alpha-\beta, \quad \psi=180^{\circ}-2 \alpha-\beta-\gamma-\varphi=90^{\circ}-\alpha-\varphi
$$

The Law of Sines for triangles $D A K$ and $E A K$ gives

$$
\frac{A K}{\sin \beta}=\frac{D K}{\sin \varphi}, \quad \frac{A K}{\sin \gamma}=\frac{E K}{\sin \psi}
$$

Since $D K=E K$, we can eliminate the sides from these equations. We get

$$
\begin{aligned}
& \sin \gamma \sin \varphi=\sin \beta \sin \psi \Longleftrightarrow \cos (\alpha+\beta) \sin \varphi=\sin \beta \cos (\alpha+\varphi) \Longleftrightarrow \\
& \Longleftrightarrow \sin (\alpha+\beta+\varphi)+\sin (\varphi-\alpha-\beta)=\sin (\alpha+\beta+\varphi)+\sin (\beta-\alpha-\varphi) \Longleftrightarrow 2 \sin (\varphi-\beta) \cos \alpha=0
\end{aligned}
$$

Then

$$
\angle D A K=\varphi=\beta=90^{\circ}-\alpha-\gamma=90^{\circ}-\angle C B D
$$

from which $A K \perp B C$.