3A. The altitudes of the acute-angled triangle $ABC$ intersect at point $O$, such that $\overline{OC} = \overline{AB}$. Determine the angle $ACB$?
Solution. Let the foot of the altitude from $B$ be denoted by $D$, and that from $C$ by $E$. Then triangle $A B D$ is congruent to triangle $O C D$ (right-angled, $\overline{O C}=\overline{A B}$ from the condition of the problem, and $\measuredangle A B D=$ $\measuredangle A C E$ as angles with perpendicular sides). From this, it follows that $\overline{B D}=\overline{C D}$, so triangle $B C D$ is an isosceles right-

![](https://cdn.mathpix.com/cropped/2024_06_05_e2d2e4c7452aa39480e3g-01.jpg?height=312&width=368&top_left_y=1887&top_left_x=1114)
triangle. Thus, $\measuredangle A C B=\measuredangle D C B=\measuredangle C B D=45^{\circ}$.