8,9

In triangle $ABC$, the altitude $AM$ is not less than $BC$, and the altitude $BH$ is not less than $AC$. Find the angles of triangle $ABC$.
Given that $B H \geq A C$, and since the perpendicular is shorter than the oblique, then $B H \geq A C \geq A M$. Similarly, $A M \geq B C \geq B H$. 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 $45,45,90^{\circ}$.