USA IMO 2003

Extend segment $DM$ through $M$ to $G$ such that $FG\parallel CD$. Then $MF=MC$ if and only if quadrilateral $CDFG$ is a parallelogram, or, $FD\parallel CG$. Hence $MC=MF$ if and only if $\angle GCD=\angle FDA$, that is, $\angle FDA+\angle CGF=180^{\circ }$.

Because quadrilateral $ABED$ is cyclic, $\angle FDA=\angle ABE$. It follows that $MC=MF$ if and only if $$180^{\circ }=\angle FDA+\angle CGF=\angle ABE+\angle CGF,$$ that is, quadrilateral $CBFG$ is cyclic, which is equivalent to $$\angle CBM=\angle CBG=\angle CFG=\angle DCF=\angle DCM.$$ Because $\angle DMC=\angle CMB$, $\angle CBM=\angle DCM$ if and only if triangles $BCM$ and $CDM$ are similar, that is $$\frac{CM}{BM}=\frac{DM}{CM},$$ or $MB\cdot MD=M{C}^2$.

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Alternative solution.

We first assume that $MB\cdot MD=M{C}^2$. Because $\frac{MC}{MD}=\frac{MB}{MC}$ and $\angle CMD=\angle BMC$, triangles $CMD$ and $BMC$ are similar. Consequently, $\angle MCD=\angle MBC$. Because quadrilateral $ABED$ is cyclic, $\angle DAE=\angle DBE$. Hence $$\angle FCA=\angle MCD=\angle MBC=\angle DBE=\angle DAE=\angle CAE,$$ implying that $AE\parallel CF$, and so $\angle AEF=\angle CFE$. Because quadrilateral $ABED$ is cyclic, $\angle ABD=\angle AED$. Hence $$\angle FBM=\angle ABD=\angle AED=\angle AEF=\angle CFE=\angle MFD.$$ Because $\angle FBM=\angle DFM$ and $\angle FMB=\angle DMF$, triangles $BFM$ and $FDM$ are similar. Consequently, $\frac{FM}{DM}=\frac{BM}{FM}$, or $F{M}^2=BM\cdot DM=C{M}^2$. Therefore $M{C}^2=MB\cdot MD$ implies $MC=MF$.

Now we assume that $MC=MF$. Applying Ceva's Theorem to triangle $BCF$ and cevians $BM,CA,FE$ gives $$\frac{BA}{AF}\cdot \frac{FM}{MC}\cdot \frac{CE}{EB}=1,$$ implying that $\frac{BA}{AF}=\frac{BE}{EC}$, so $AE\parallel CF$. Thus, $\angle DCM=\angle DAE$. Because quadrilateral $ABED$ is cyclic, $\angle DAE=\angle DBE$. Hence $$\angle DCM=\angle DAE=\angle DBE=\angle CBM.$$ Because $\angle CBM=\angle DCM$ and $\angle CMB=\angle DMC$, triangles $BCM$ and $CDM$ are similar. Consequently, $\frac{CM}{DM}=\frac{BM}{CM}$, or $C{M}^2=BM\cdot DM$.