3. Given an acute-angled triangle $A B C$ with angle $\angle A B C=\alpha$. On the extension of side $В C$ a point $D$ is taken such that the line $A D$ is tangent to the circumcircle $\omega$ of triangle $A B C$. The line $A C$ intersects the circumcircle of triangle $A B D$ at point $E$. It turns out that the bisector of $\angle A D E$ is tangent to the circle $\omega$. In what ratio does point $C$ divide the segment $A E$?
Answer: $A C: C E=\sin \alpha$.

Solution. Let $r$ be the radius of $\omega$, and $K$ and $L$ be the points of intersection of the angle bisector of $\angle A D E$ with $A E$ and $\omega$ respectively (see figure). The angle between the tangent $A D$ to the circle $\omega$ and the chord $A C$ is equal to the inscribed angle that subtends $A C$, hence $\angle D A E = \angle A B D$. Moreover, the inscribed angles $\angle A B D$ and $\angle A E D$ subtend the chord $A D$ and are therefore equal. Thus, $\angle D A E = \angle A E D$, which means that triangle $A D E$ is isosceles.

Therefore, $A K = K E$ and $D K \perp A E$. The segment $A E$ is parallel to the radius $O L$ of the circle $\omega$, since both are perpendicular to $D L$. Then $A C = 2 r \sin \alpha$ and

$$
C E = A E - A C = 2 A K - A C = 2(r + r \sin \alpha) - 2 r \sin \alpha = 2 r
$$

which gives the answer.

![](https://cdn.mathpix.com/cropped/2024_05_06_9304f7a3bc6f36cb1f9ag-18.jpg?height=502&width=528&top_left_y=517&top_left_x=821)