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.
 A: Position the circumcenter $P$ of the triangle at the origin, and let the vectors from the $P$ to $A$, $B$, and $C$ be $\vec{A}$, $\vec{B}$, and $\vec{C}$. Then the orthocenter is at $\vec{A}+\vec{B}+\vec{C}$. (Proof: the vector from $A$ to this point is $(\vec{A}+\vec{B}+\vec{C})-\vec{A} = \vec{B}+\vec{C}$. The vector coinciding with the side opposite vertex $A$ is $\vec{B}-\vec{C}$. Now $(\vec{B}+\vec{C})\cdot(\vec{B}-\vec{C}) = |\vec{B}|^2 - |\vec{C}|^2 = R^2-R^2 = 0$, where $R$ is the circumradius. So the line through $A$ and the head of $\vec{A}+\vec{B}+\vec{C}$ is the altitude to $BC$. Similarly for the other three altitudes.)
Now the vector coinciding with $OC$ is $\vec{O}-\vec{C}=\vec{A}+\vec{B}$. Thus $|OC|=|AB|$ if and only if
$$|\vec{A}+\vec{B}|^2 = |\vec{A}-\vec{B}|^2$$
if and only if
$$\vec{A}\cdot\vec{A} + \vec{B}\cdot\vec{B} + 2\vec{A}\cdot\vec{B} = \vec{A}\cdot\vec{A} + \vec{B}\cdot\vec{B} - 2\vec{A}\cdot\vec{B}$$
if and only if
$$4\vec{A}\cdot\vec{B} = 0$$
if and only if
$$\angle APB = \pi /2$$
if and only if
$$\boxed{\angle ACB = \pi/4 = 45^\circ}.$$
A: 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
$$\begin{eqnarray}
\sin B &=& \sin C \; (\cos A+\sin A) \\
\implies\sin(A+C) &=& \cos A \sin C + \sin A \sin C \\
\implies\sin A \cos C + \cos A \sin C &=& \cos A \sin C + \sin A \sin C \\
\implies\sin A \cos C &=& \sin A \sin C \\
\implies\cos C &=& \sin C \\
\implies C &=& \pi/4
\end{eqnarray}$$
A: 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. 
