10.10. In triangle $A B C$, the altitude $A M$ is not less than $B C$, and the altitude $B H$ is not less than $A C$. Find the angles of triangle $A B C$.
10.10. According to the condition $B H \geqslant A C$, and since the perpendicular is shorter than the slant, then $B H \geqslant A C \geqslant A M$. Similarly $A M \geqslant B C \geqslant B H$. Therefore $B H=A M=A C=$ $=B C$. Since $A C=A M$, the segments $A C$ and $A M$ coincide, i.e., $\angle C=90^{\circ}$, and since $A C=B C$, the angles of triangle $A B C$ are $45,45,90^{\circ}$.