3. On the extension of side $B C$ of triangle $A B C$, 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 $E$, and it is given that $A C: C E=1: 2$. It turns out that the bisector of angle $A D E$ is tangent to the circle $\omega$. Find the angles of triangle $A B C$.
Answer: $30^{\circ}, 60^{\circ}, 90^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_9304f7a3bc6f36cb1f9ag-07.jpg?height=554&width=580&top_left_y=494&top_left_x=795)

Solution 1. Let $M$ be the midpoint of side $A C$, and $K$ 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 $\angle D A E = \angle A B D$. 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 triangle $A D E$ is isosceles. Therefore, the bisector $D K$ of triangle $A D E$ is also its median and altitude. In particular, $D K \perp A E$, so $O L \| A E$. Then

$$
A O = O L = M K = 2 A M, \quad \text{from which } \angle A B C = \angle A O M = 30^{\circ}.
$$

Thus, $\angle A D K = 60^{\circ}$, and the isosceles triangle $A D L$ turns out to be equilateral. Therefore,

$$
\angle L B C = \angle L A C = \angle L A D - \angle K A D = 30^{\circ} = \angle A B C
$$

which means $B D$ is the angle bisector of $\angle A B L$. Thus, $C$ is the midpoint of the arc $A L$, so $D B$ is the angle bisector of $\angle A D L$. Therefore, $\angle K D B = 30^{\circ}$. Then $\angle A C B = 60^{\circ}$, and the remaining angle in the triangle is $90^{\circ}$.

Solution 2. Let $\alpha = \angle A B C$, $r$ 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. 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 triangle $A D E$ is isosceles. Therefore, the bisector $D K$ of triangle $A D E$ is also its median and altitude. In particular, $D K \perp A E$, so $O L \| A E$. Then

$$
r \sin \alpha + r = A K = K E = \frac{1}{2} A E = \frac{3}{2} A C = 3 r \sin \alpha, \quad \text{from which } \quad \sin \alpha = \frac{1}{2}
$$

Note that $\alpha < \angle A B L = \frac{1}{2} \angle A O L < \frac{\pi}{2}$, since quadrilateral $D A O L$ is cyclic. Therefore, $\alpha = \frac{\pi}{6}$. Moreover,

$$
C K = A K - A C = A K - \frac{2}{3} A K = \frac{1}{3} K E
$$

from which

$$
\operatorname{tg} \angle A C B = \operatorname{tg} \angle D C K = \frac{D K}{C K} = 3 \cdot \frac{D K}{K E} = 3 \operatorname{tg} \alpha = \sqrt{3}
$$

Therefore, $\angle A C B = \frac{\pi}{3}$ and $\angle B A C = \pi - \alpha - \angle A C B = \frac{\pi}{2}$.