[ Isosceles, Inscribed, and Circumscribed Trapezoids ]

[Properties and characteristics of isosceles triangles. ]

Let $M$ be the point of intersection of the diagonals of a convex quadrilateral $ABCD$, in which sides $AB$, $AD$, and $BC$ are equal to each other.

Find the angle $CMD$, given that $DM = MC$, and $\angle CAB \neq \angle DBA$.
Let $\angle A B D=\angle A D B=\alpha, \angle B A C=\angle A C B=\beta$. By the external angle theorem of a triangle, $\angle B M C=\alpha+\beta$. Draw a line through point $A$ parallel to side $C D$. Let this line intersect line $D B$ at point $K$. Triangle $AMK$ is isosceles because it is similar to the isosceles triangle $CMD$. Therefore, $\angle D K=D M+M K=C M+M A=C A$, which means trapezoid $A K C D$ is isosceles. Thus, $C K=A D=B C$, so triangle $B C K$ is also isosceles (by the condition that point $K$ does not coincide with point $B$). Additionally, $\angle K C M=\angle A D M=\alpha$. Consider two cases.

1) Point $K$ lies on diagonal $D B$. Then $\angle K B C=\angle B K C=\angle K M C+\angle K C M=2 \alpha+\beta$. Therefore, $180^{\circ}=\angle B M C+\angle M B C+\angle M C B=(\alpha+\beta)+(2 \alpha+\beta)+\beta=3 \alpha+3 \beta$.
2) Point $K$ lies on the extension of $D B$ beyond point $B$. Then $\angle B K C=\angle K B C=\angle B M C+\angle B C M=\alpha+2 \beta$. Therefore, $180^{\circ}=\angle K M C+\angle M K C+\angle K C M=(\alpha+\beta)+(\alpha+2 \beta)+\alpha=3 \alpha+3 \beta$.

Thus, in either case, $\alpha+\beta=60^{\circ}$. Consequently, $\angle C M D=180^{\circ}-\angle K M C=180^{\circ}-(\alpha+\beta)=120^{\circ}$.

## Answer

$120^{\circ}$.