We will prove $BF \bot MM'$ which will obviously solve the problem. Take $H$ to be the projection of $B$ on $MM'$. Now, it is well-known (theorem of the 6 points that $P,Q,R,P',Q',R'$ lie on a circle centered in $X$, the midpoint of $MM'$). Now look at the circles $(PQR), (BM), (BM')$ (where $(XY)$ stands for the circle with diameter $XY$ and $(XYZ)$ stands for the circumcircle of $\triangle{XYZ}$). Their radical axes are $RP, R'P', BH$ which are concurrent, so $BH$ passes through $F$, done.

Dear Mathlinkers, you can see : http://perso.orange.fr/jl.ayme vol. 2 Les deux points de Schroeter p. 4-6 and adapt the same thechnic for two isogonal points... as it is mentioned p. 6 in the commentar. Sincerely Jean-Louis

Note that $\angle{CQ'P'}=\angle{CM'P'}=90-\angle{M'CP'}=90-\angle{MCQ}=\angle{QMC}=\angle{QPC}$ so $QQ'P'P$ is cyclic.The radical axis of $QMPC$ and $QQ'P'P$ is $PQ$,while that of $Q'MP'C$ and $QQ'P'P$ is $P'Q'$.Hence $G=PQ \cap P'Q'$ is a point on the radical axis of $PMQC$ and $P'M'Q'C$.Hence $CG$ is the radical axis of the two circles so the centre joining line is perpendicular to $CG$.From this it follows that $MM' \perp CG$.Analogously $AE \perp MM'$ and $BF \perp MM'$ so the conclusion follows.