Let the parallel to $CA$ through $E$ cut $AB$ at $G'.$ From $\triangle BEG' \sim \triangle BCA$ we get $\frac{CE}{CB}=\frac{AG'}{AB}=1-\frac{CF}{CA} \Longrightarrow \frac{CF}{CA}=\frac{AB-AG'}{BA}=\frac{BG'}{BA} \Longrightarrow FG' \parallel CB$ Thus, $CEG'F$ is a parallelogram $\Longrightarrow$ $CM$ passes through $G'$ $\Longrightarrow$ $G \equiv G'.$ $\angle EGF=\angle BCA,$ $\angle CEF=\angle EFG=\angle CAB$ imply that $\triangle ABC \sim \triangle FEG.$

Since $\triangle CFE \sim \triangle ABC \implies CE \cdot CB = CF \cdot CA $ $(1)$. From hypotesis we have that $CE \cdot CA = CB \cdot FA$ $ (2)$. Multiplying $(1)$ and $(2)$ we get $CE^2=CF \cdot CA$ $(3)$. Now we have that $CG$ is symmedian, so $\frac{AG}{GB}=\frac{AC^2}{BC^2}=\frac{CE^2}{CF^2}$ $ (4)$. Finally from $(3)$ aqnd $(4)$ we get $\frac{FA}{CF}=\frac{AG}{GB} \implies FG \parallel BC$ analogously $GE \parallel AC \implies \triangle FGE \sim \triangle CFE \sim \triangle ABC$

Note that $\frac{CE}{CB}+\frac{CF}{AC}=1 \implies \frac{CE}{CB}=\frac{AF}{AC}=k \implies$ parallels through $F$ and $E$ to $BC$ and $AC$ meet at a common point $G'$.Now let the $C$-median of $\triangle{CEF}$ meet $AB$ at $G$.As $AFEB$ is cyclic it follows that $CG$ is the $C$-symmedian of $\triangle{ABC} \implies \frac{AG}{BG}=\frac{b^2}{a^2}$.Also $CE*CB=CF*CA \implies ka^2=(1-k)b^2 \implies \frac{k}{1-k}=\frac{b^2}{a^2}$.Therefore $\frac{AG'}{BG'}=\frac{CE}{BE}=\frac{k}{1-k}=\frac{b^2}{a^2}=\frac{AG}{BG} \implies G \equiv G'$ and the result immmediately follows.