Problem: Let $ABC$ be a triangle.
A circle passing through $A$ and $B$
intersects segments $AC$ and $BC$ at $D$ and $E$, respectively.
Lines $AB$ and $DE$ intersect at $F$,
while lines $BD$ and $CF$ intersect at $M$.
Prove that $MF = MC$ if and only if $MB \cdot MD = MC^2$.

Solution: Ceva theorem plus the similar triangles.

\begin{center}
\begin{asy}
  size(7.5cm);
  pair C = Drawing("C", (3,0), dir(-45));
  pair D = Drawing("D", (0,0));
  pair M = Drawing("M", (1,-0.7));
  pair F = Drawing("F", 2*M-C);
  pair B = Drawing("B", conj(C/M) * (M-C)+C, dir(130));
  pair E = Drawing("E", extension(D,F,B,C), dir(40));
  pair A = Drawing("A", extension(D,C,B,F), dir(225));
  draw(A--B--C--cycle);
  draw(circumcircle(B,D,E), dashed);
  draw(A--F, dotted);
  draw(E--F--C, dotted);
  draw(B--M, dotted);
\end{asy}
\end{center}

We know unconditionally that
\[ \dang CBD = \dang EBD = \dang EAD = \dang EAC. \]
Moreover, by Ceva's theorem on $\triangle BCF$,
we have $MF = MC \iff \ol{FC} \parallel \ol{AE}$.
So we have the equivalences
\begin{align*}
  MF = MC &\iff \ol{FC} \parallel \ol{AE} \\
  &\iff \dang FCA = \dang EAC \\
  &\iff \dang MCD = \dang CBD \\
  &\iff MC^2 = MB \cdot MD.
\end{align*}