Answer: $\angle A C M=45^{\circ}$. Solution. Since triangle $B A K$ is a right isosceles triangle, $\angle A K B=45^{\circ}$. Let the bisector of angle $C A D$ intersect segment $B K$ at point $N$. Triangles $A N K$ and $A N C$ are equal: $A N$ is common, $A C=A K, \angle C A N=\angle K A N$. Therefore, $\angle N C A=\angle N K A=45^{\circ}$. Therefore, $C N$ is the bisector of the right angle $A C D$, and $N$ is the point of intersection of the bisectors of triangle $A C D$. Thus, point $N$ lies on the bisector of angle $A C D$ and on segment $B K$, that is, it coincides with point $M$. Therefore, $\angle A C M=\angle A C N=45^{\circ}$.
3. In quadrilateral $A B C D$, side $A B$ is equal to diagonal $A C$ and is perpendicular to side $A D$, and diagonal $A C$ is perpendicular to side $C D$. A point $K$ is taken on side $A D$ such that $A C=A K$. The bisector of angle $A D C$ intersects $B K$ at point $M$. Find the angle $A C M$. (R. Zhenodarov)