3. In triangle $ABC$, the median $AM$ is drawn. Circle $\alpha$ passes through point $A$, touches line $BC$ at point $M$, and intersects sides $AB$ and $AC$ at points $D$ and $E$ respectively. On the arc $AD$ that does not contain point $E$, a point $F$ is chosen such that $\angle BFE = 72^{\circ}$. It turns out that $\angle DEF = \angle ABC$. Find the angle $\angle CME$.
Answer: $36^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_9304f7a3bc6f36cb1f9ag-33.jpg?height=445&width=551&top_left_y=1068&top_left_x=844)

Solution. Inscribed angles $D A F$ and $D E F$ subtend the arc $D F$ and are therefore equal. Given that $\angle F A B = \angle A B C$, it follows that $B C \parallel A F$. Then the line $\ell$, passing through point $M$ and perpendicular to the tangent $B C$, contains the diameter of the circle $\omega$ and is perpendicular to its chord $A F$. Thus, points $A$ and $F$ are symmetric with respect to $\ell$, meaning that $\ell$ is the axis of symmetry of the trapezoid $B C A F$. This implies that $A C = B F$ and $A M = F M$. Therefore, triangles $B F M$ and $C A M$ are congruent by the three sides, so $\angle B F M = \angle E A M$. Additionally, $\angle E F M = \angle E A M$ as inscribed angles subtending the same chord. Thus, $F M$ is the angle bisector of $\angle B F E$, and

$$
\angle E A M = \angle E F M = \frac{1}{2} \angle B F E = 36^{\circ}.
$$

Notice now that the angle between the chord $E M$ and the tangent $B C$ is equal to the inscribed angle subtending $E M$. Therefore, $\angle C M E = \angle E A M = 36^{\circ}$.