Let the equal angles be $x$. Then $BAM$ is isosceles and $\angle B=90-x,$ and therefore $\angle C=90-2x$. Thus $0<x<45^\circ$. Now $(BAM)=(MAC)$ and thus $\frac{1}{2}\cdot c\cdot AM\cdot \sin (2x)=\frac{1}{2}\cdot b\cdot AM\cdot \sin x$; therefore $b=2c\sin x$. But from the law of sines we have $\frac{b}{\cos x}=\frac{c}{\cos 2x}$. As a result, $\cos 2x=\frac{1}{2}$ and therefore $x=30^\circ.$ The angles of the triangle are: $\angle A=90^\circ, \angle B=60^\circ, \angle C=30^\circ.$

In the triangle $ABC$, $D$ is the foot of the altitude from $A$ to $BC$, and $M$ is the midpoint of the line segment $BC$. The three angles $\angle BAD$, $\angle DAM$ and $\angle MAC$ are all equal. Find the angles of the triangle $ABC$.