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Dear Mathlinlers, an outlook of my synthetic proof 1. EF is the polar of B (EF perpendicular to AB) 2. X the point of intersection of EF and AB is the pole of B 3. the circle with diameter AB goes through E and F 4. according to the three chords theorem, CD goes through X 5. according to the de La Hire theorem, M is on the polar of X (BM perpendicular to AB) 6. according to a converse of the Reim’s theorem (EF // MB), we are done… Sincerely Jean-Louis

Let $AO$ intersect $c$ again at $P$. Now $PC \perp CA \wedge PD \perp DA \implies PC$ and $PD$ touches $c_1$. And let $FC \cap ED = Q$. Applying Pascal on $CCEDDF$, we get $P, M, Q$ collinear and applying Pascal on $EECFFD$, we get $B, M, Q$ collinear. So, $P, B, M, Q$ are all collinear. And $PA$ is the diameter of $c \implies AB \perp QM$. $\therefore$ we must have $B$ is the Miquel point of the cyclic quad $CEDF$. Thus $B \in \odot MCF$. Done!

Let the circle $BEAF$ be $c_2$, then inversion wrt $ \odot (A, AC)$ implies- \[B = c \cap c_2 \Rightarrow \text{ (inverse of } B ) = CD \cap EF\] As wrt $\odot (A,AC)$, inverse of $c = CD$ and inverse of $c_2 = EF$ So, $B$ is the miquel point of cyclic quad $CEDF$, and thus $ B\in\odot MCF $.

I thonk I have a nice solution : Let Q be the point where EF meet CD. Aplying Pascal's Theorem to EEDFC, we get that T, B, M are collinear , where T is the point where ED meets CF. Moreover, in cyclic quadrilateral ECFD, T, M are the points where opposite sides meet and Q the point where its diagonals meets, so T is the polar of Q, so TM is vertical to AQ. So: EF||TM, since AB is vertical to EF. Finally: ^BMC=^MEF=^CFB, and BCF is cyclic.