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.

Assuming there is an answer, then it is $45^\circ$ or $\pi/4$, as a symmetric right-angled triangle (half a square), where $C$ is not the right angle, satisfies this: the orthocentre is at the right angle. 
