3. The altitudes $C M, A P$ in the acute-angled triangle $A B C$ intersect at point $H$. Determine the angle $A C B$, if $\overline{A B}=\overline{C H}$.
Solution. Let's consider triangles $A B P$ and $C H P$ (diagram on the right). For them, $\overline{A B}=\overline{C H}$. Then the angles at vertex $P$ are right angles, and $\measuredangle B A P=\measuredangle H C P$, as angles with mutually perpendicular sides. Therefore, $\measuredangle A B P=\measuredangle C H P$. Now, by the ASA criterion, it follows that $\triangle A B P \cong \triangle C H P$, from which we get $\overline{A P}=\overline{C P}$. Thus, $\triangle A P C$ is isosceles right-angled, so $\measuredangle A C B=\measuredangle A C P=45^{\circ}$.

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