Circle through the circumcentre of a triangle problem

Question

Let ABC be an acute triangle and O it's circumcentre. Let S denote the circle through A,B, O. The lines CA and CB meet S again at P and Q, respectively.

Prove that the lines CO and PQ are perpendicular.

My initial idea was to construct the triangle so that the circumcentre was the origin of the 2d plane, and then use vector equation of the lines CO and PQ and the scalar product to show that they are perpendicular, however I haven't been able to find the vector equation of PQ and am really struggling.

Any ideas?

Answer 1

Hint: Let the intersection of $CO,PQ$ be $R$. Letting angles of the triangle be $\alpha,\beta,\gamma$ find values of angles $RPC$ and $RCP$.

Answer 2

1) Will the green marked angles equal to each other?

2) If yes, the red marked angle should be equal to each other. Right?

3) What is the size of $\angle QBC’$?

Answer 3

I will follow Lucian's hint and show that $O$ is the orthocentre of triangle $CPQ$.

It is enough to show that $PO$ is perpendicular to $CB$ and that $QO$ is perpendicular to $CA$. We will prove the first of these statements, and the other will follow by symmetry.

Since $OB = OA$, the arcs $OA$ and $OB$ are opposite. Hence by the inscribed angle theorem, the oriented angles $\angle OPA$ and $\angle OPB$ are opposite modulo $\pi$. (This is only true modulo $\pi$ because $P$ and $O$ could be on the same side of $AB$.)

Thus if $r$ is the reflection about line $PO$, the image of line $PB$ under $r$ is line $PA$. Since $PO$ is a diameter of circle $ABC$, the circle is invariant under $r$. Therefore $r(B)$ must belong both to circle $ABC$ and to line $PA$, hence $r(B)$ is either $A$ or $C$. It is easy to see that $r(B) \ne A$, since otherwise $PO$ would be a diameter of circle $OAB$ and, consequently, $PA$ would be tangent to circle $ABC$. Thus $r(B) = C$, which proves that $PO$ is perpendicular to $BC$.