4. A right triangle $A B C$ (cathetus $B C$ is greater than cathetus $A C$) is inscribed in a circle. A point $D$ is chosen on side $B C$ such that $B D=A C$, and point $M$ is the midpoint of the arc $A C B$. Find the angle $C D M$.
Answer: $45^{\circ}$.

Solution. Note that $A B$ is the diameter of the circumscribed circle. Connect point $M$ with points $A, B, C$, and $D$. Since arcs $A M$ and $B M$ are equal, the chords $A M$ and $B M$ that subtend them are also equal. Segments $B D$ and $A C$ are equal by the condition. Finally, angles $M B C$ and $M A C$ are equal, as inscribed angles subtending the same arc. Thus, triangles $M B D$ and $M A C$ are equal by two sides and the angle between them. This implies the equality of sides $M D$ and $M C$, meaning that triangle $M C D$ is isosceles. It remains to note that $\angle M C D=\angle M C B=\angle M A B=45^{\circ}$, since arc $M B$ is half of the circle. Criteria for evaluation. Proof of the equality of triangles $M B D$ and $M A C$ - 3 points.