265. Inside the square $A B C D$, an equilateral triangle $A B M$ is constructed, and vertex $M$ is connected to points $C$ and $D$. Find the angles of triangle $C M D$.
$\Delta$ The angles of an equilateral triangle are $60^{\circ}$, so the angle at vertex $A$ of the isosceles triangle $A M D$ is $90-60=30^{\circ}$. Therefore, the angles at its base are $(180-30) / 2=75^{\circ}$. Thus, the angle $C D M$ is $90-75=15^{\circ}$ (as well as $\angle M C D$ ). $\triangleleft$