Let AB be a diameter of a circle C, M a point on the circle, K a point on the Diameter. If P and Q are the circumcenters of the triangles AMK and BMK, show that the KPMQ is cyclic.
I have tried for quite a while now, but all I can seem to get is that PQ is perpendicular to MK and that if it were cyclic, then the angles OMQ and OKQ must be of 90 degrees. But I Can’t show any of that.
Any insight, please?
Your approach is correct. Let $PQ$ meet $MK$ at T. Now notice that $PQ$ is the perpendicular bisector of $MK$. Also, since $p$ and $Q$ are circumcentres of the respective triangles, notice that $\Delta PMK$ and $\Delta QMK$ are isoceles. Now since $PQ$ is the perpendicular bisector of $MK$, $PQ$ will also bisect $\angle MPK$ and $\angle MQK$. Hence,
$$\angle TPK = \angle TPM = \angle KPM /2 = \angle KAM$$
The last equality also comes from the fact that $P$ is the circumcentre of $\Delta KAM$. Similarly,
$$\angle TQK = \angle TQM = \angle KQM /2 = \angle KBM$$
From these two equations, we get
$$\angle TQK + \angle TPK= \angle TQM + \angle TPM = \angle KBM + \angle KAM = 90^\circ$$
The last equality here is due to the fact that $\Delta AMB$ is right angled.
This shows that $\Delta PQK$ and $\Delta PMK$ are right angled and hence that $KPMQ$ is cyclic, in fact, with diameter $PQ$.