8.3. Given an acute-angled triangle $A B C$. Point $M$ is the intersection point of its altitudes. Find the angle $A$, if it is known that $A M=B C$.

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83. Given an acute-angled triangle $A B C$. Point $M$ is the intersection point of its altitudes. Find the angle $A$, if it is known that $A M=B C$.

Answer: $45^{\circ}$. Hint Let $\mathrm{K}$ be the foot of the altitude from point В. We will prove that triangles АМ К and BKC are equal. Indeed, we have right triangles, in which $\angle \mathrm{MAK}=\angle \mathrm{CBK}=90^{\circ}-\angle \mathrm{C}$ and, by condition, $A M=B C$. Then, from the equality of the triangles, it follows that $A K=B K$, and therefore, in the right triangle $A B K$, the legs are equal. Therefore, $\angle A=45^{\circ}$.