Find the angle in a triangle if the distance between one vertex and orthocenter equals the length of the opposite side Let $O$ be the orthocenter (intersection of heights) of the triangle $ABC$. If $\overline{OC}$ equals $\overline{AB}$, find the angle $\angle$ACB.

Let point $P$ on $AC$ be the foot of the perpendicular $BO$, and note that $\angle OCA$ is the complement of $A$. Then,
$$\begin{eqnarray}
|AC|&=&|AP|+|PC|\\
&=&|AB|\cos A+|OC|\cos\angle OCA \\
&=&|AB| \cos A+|OC| \sin A \\
&=&|AB|(\cos A+\sin A)
\end{eqnarray}$$
Conveniently scaling to unit circumdiameter ---so that $|AC| = \sin B$, $|AB| = \sin C$, and $|BC| = \sin A$ (which we may assume is non-zero)--- we have
$$\begin{eqnarray}
\sin B &=& \sin C \; (\cos A+\sin A) \\
\implies\sin(A+C) &=& \cos A \sin C + \sin A \sin C \\
\implies\sin A \cos C + \cos A \sin C &=& \cos A \sin C + \sin A \sin C \\
\implies\sin A \cos C &=& \sin A \sin C \\
\implies\cos C &=& \sin C \\
\implies C &=& \pi/4
\end{eqnarray}$$
