$ABC$ is similar to $QPC$,Angle-chasing make it easy.

Can anyone think of a proof using inversion. It calls for it

Then $\angle of intersection =180-(90-\angle C+\angle C)=90$ Thus $PQ \perp AO$

Dear Mathlinkers, according to the Reim's theorem the tangent to (O) at C is parallel to PQ...and we are done Sincerely Jean-Louis

Inversion proof. Carry out an inversion centre $C$ with radius, $r$, satisfying $r^{2} = CA \cdot CP = CB \cdot CQ$ (intersecting chords) -- this will map $A \leftrightarrow P$ and $B \leftrightarrow Q$. $PQ$ is a line not through $O$ so will map to a circle through $C$, the image of $P$ and the image of $Q$ i.e. $PQ \leftrightarrow \circ ABC$. $OC$ is a line through $C$ so will map to itself. As angles preserved by inversion it suffices to show that $OC$ is perpendicular to circle $ABC$ -- this follows from $OC$ being a radius of circle $ABC$.

frill wrote: Inversion proof. Carry out an inversion centre $C$ with radius, $r$, satisfying $r^{2} = CA \cdot CP = CB \cdot CQ$ (intersecting chords) -- this will map $A \leftrightarrow P$ and $B \leftrightarrow Q$. $PQ$ is a line not through $O$ so will map to a circle through $C$, the image of $P$ and the image of $Q$ i.e. $PQ \leftrightarrow \circ ABC$. $OC$ is a line through $C$ so will map to itself. As angles preserved by inversion it suffices to show that $OC$ is perpendicular to circle $ABC$ -- this follows from $OC$ being a radius of circle $ABC$. Yeah Just inversion.but need a proof of inversion at Exam

$\angle CPQ= \angle B$, $\angle OCP=\angle OCA=90-\angle B$ so $\angle CPQ+\angle OCP=90$

galav wrote: Can anyone think of a proof using inversion. It calls for it Well.... it doesn't exactly call for it because it is pretty easy, so I think that inverting is just an overkill.

$\triangle CPQ$ and $\triangle CAB$ are inversley similar, and then by isogonal conjugates $OC$ must coincide with the altidude from $C$ to $PQ$

Let $OC$ and $PQ$ intersect at M $\measuredangle BAO=\measuredangle OBA=\alpha$ $\measuredangle CBO=\measuredangle OCB=\beta$ $\measuredangle ACO=\measuredangle OAC=\gamma$ $\measuredangle BAP=\measuredangle BAC=\measuredangle BAO+\measuredangle OAC=\alpha+\gamma$ Look at $\triangle ABC$. $2\alpha+2\beta+2\gamma=180$ $ABQP$ is cyclic quadrilateral,so $\measuredangle BQP=\measuredangle BAP=\alpha+\gamma$ $B Q C$ colinear $\Rightarrow \measuredangle BQP=\measuredangle CQP=\alpha+\gamma$ and $\measuredangle OCB=\measuredangle OCQ=\beta$ $Q P M$ colinear $\Rightarrow \measuredangle CQP=\measuredangle CQM=\alpha+\gamma$ $O C M$ colinear $\Rightarrow \measuredangle OCQ= \measuredangle MCQ=\beta$ Look at $\triangle CQM.$ $\measuredangle CQM+\measuredangle QMC+ \measuredangle MCQ=180$ and $\measuredangle CQM+ \measuredangle MCQ=\alpha+\beta+\gamma=90$, so $\measuredangle QMC=90$

Draw the tangent to $(ABC)$ at $C$ it will be perpendicular to $OC$.It's only left to prove that the tangent line and $PQ$ are paralell which can be proven by Reim's theorem

This is due to the fact that the circumcentre and the orthocentre are isogonal conjuagtes.