3. Point $M$ is the midpoint of side $AB$ of triangle $ABC$. A circle $\omega_{1}$ is drawn through points $A$ and $M$, tangent to line $AC$, and a circle $\omega_{2}$ is drawn through points $B$ and $M$, tangent to line $BC$. Circles $\omega_{1}$ and $\omega_{2}$ intersect again at point $D$. Point $E$ lies inside triangle $ABC$ and is symmetric to point $D$ with respect to line $AB$. Find the angle $CEM$.
Answer: $180^{\circ}$
![](https://cdn.mathpix.com/cropped/2024_05_06_627b343ad954e1e3a31bg-23.jpg?height=468&width=1264&top_left_y=1985&top_left_x=424)

Solution 1. The angle between the tangent $A C$ and the chord $A M$ of circle $\omega_{1}$ is equal to the inscribed angle in $\omega_{1}$ that subtends $A M$, i.e., $\angle C A B = \angle A D M$. Similar reasoning for $\omega_{2}$ gives $\angle A B C = \angle B D M$. Extend triangle $A D B$ to parallelogram $A D B K$ (see the right figure). Triangles $A C B$ and $D A K$ are similar by two angles. Point $M$ is the midpoint of segments $A B$ and $D K$, so triangles $A C M$ and $D A M$ are also similar (by angle and proportional sides). Therefore, $\angle A M C = \angle A M D$. Moreover, from the symmetry of points $D$ and $E$ with respect to $A B$, it follows that $\angle A M E = \angle A M D$. Thus, point $E$ lies on segment $C M$.

Solution 2. By the properties of tangents, $\angle A D M = \angle C A M$ and $\angle M D B = \angle C B M$. Therefore,

$$
\angle A D B = \angle A D M + \angle M D B = \angle C A M + \angle C B M = 180^{\circ} - \angle A C B
$$

and quadrilateral $A C B D$ is cyclic. Then

$$
\angle B D C = \angle B A C = \angle M D A \quad \text{and} \quad \angle B C D = \angle B A D = \angle M A D.
$$

Thus, triangles $B D C$ and $M D A$ are similar by two angles, so

$$
\frac{B D}{M D} = \frac{B C}{M A} = \frac{C B}{M B}
$$

Therefore, triangles $B D M$ and $C B M$ are similar by angle and ratio of sides, and in particular, angles $\angle C M B$ and $\angle D M B$ are equal. Due to the symmetry of points $D$ and $E$ with respect to $A B$, we get

$$
\angle E M B = \angle D M B = \angle C M B
$$

Thus, point $E$ lies on line $C M$, from which $\angle C E M = 180^{\circ}$.

Solution 3. Let

$$
\alpha = \angle C A B, \quad \beta = \angle A B C, \varphi = \angle A M D, \psi = \angle A M C
$$

The angle between the tangent $A C$ and the chord $A M$ of circle $\omega_{1}$ is equal to the inscribed angle in $\omega_{1}$ that subtends $A M$, i.e., $\angle C A B = \angle A D M$. Similar reasoning for $\omega_{2}$ gives $\angle A B C = \angle B D M$. Note that

$$
\angle D A M = 180^{\circ} - \alpha - \varphi \quad \angle D M B = 180^{\circ} - \beta - (180^{\circ} - \varphi) = \varphi - \beta
$$

Then the Law of Sines for triangles $A D M$ and $D M B$ gives

$$
\frac{A M}{\sin \alpha} = \frac{D M}{\sin (\alpha + \varphi)}, \quad \frac{B M}{\sin \beta} = \frac{D M}{\sin (\varphi - \beta)}
$$

Since $A M = B M$, we can eliminate the sides from these equations. We get

$$
\frac{\sin \beta}{\sin \alpha} = \frac{\sin (\varphi - \beta)}{\sin (\alpha + \varphi)}
$$

Similar reasoning for triangles $A C M$ and $C M B$ gives

$$
\frac{\sin \beta}{\sin \alpha} = \frac{\sin (\psi - \beta)}{\sin (\alpha + \psi)}
$$

Eliminating the left side of these relations, we get

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

from which $\varphi = \psi$. Moreover, from the symmetry of points $D$ and $E$ with respect to $A B$, it follows that $\angle A M E = \varphi$. Thus, point $E$ lies on segment $C M$.