3. Given a right triangle $A B C$. On the extension of the hypotenuse $B C$, a point $D$ is chosen 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: $A C: C E=1: 2$.

![](https://cdn.mathpix.com/cropped/2024_05_06_9304f7a3bc6f36cb1f9ag-03.jpg?height=508&width=514&top_left_y=777&top_left_x=834)

Solution. Let $\alpha=\angle A B D, K$ and $L$ be the points of intersection of the angle bisector of $\angle A D E$ with $A E$ and $\omega$ respectively. 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 $\angle D A E=\alpha$. Moreover, the inscribed angles $\angle A E D$ and $\alpha$ subtend the chord $A D$ and are therefore equal. Thus, $\angle D A E=\angle A E D=\alpha$ and triangle $A D E$ is isosceles. Therefore, the bisector $D K$ is also its median and altitude. This means $A B \| D L$, since lines $A B$ and $D L$ are perpendicular to $A E$. Right triangles $A O D$ and $L O D$ are equal by the leg and hypotenuse, hence $\angle A D O=\angle L D O=\angle A B D=\alpha$. From the right triangle $A D K$ we get that $3 \alpha=\frac{\pi}{2}$ and $\alpha=\frac{\pi}{6}$. Then $A D=2 D K$, and by the property of the angle bisector

$$
\frac{C K}{A C}=\frac{D K}{A D}=\frac{1}{2},
$$

from which

$$
C E=E K+C K=A K+C K=A C+2 C K=2 A C .
$$