I have found a problem in a book of H. S. Coxeter. As a hint, he says that the triangles $OPB$ anb $PRQ$ are similar, but I don´t know why. I have dealing with angles for a while, but I can`t seee the way to involve the segment $PQ$.
I'll describe the figure more precisely. Take a triangle $ABC$ and any point $O$ inside it. $OP$ and $OQ$ are perpendicular to sides $AB$ and $BC$, respectively. $PR$ is also perpendicular to $BC$. So the problem is proving that $\angle PBO=\angle RPQ$.
Any ideas?
Notes:
- Please avoid trigonometry and Cartesian plane. It is a problem of classic geometry.
- In the figure, $P$, $Q$ and $R$ lie on the sides, but they can be outside the triangle.
I think it trivially follows from the fact that $OP$ and $OQ$ are perpendiculars to $AB$ and $BC$, respectively. That is, the quadrilateral $OPBQ$ is cyclic so: $$\angle PBO = \frac\pi 2 - \angle POB = \frac\pi 2 - \angle PQB = \angle RPQ.$$