MithsApprentice wrote:

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 $MB\cdot MD = MC^2$ .

Assume $MF = MC$ .
1) Draw parallel to $AD$ from $F$ and parallel to $ED$ from $C$ . Let them intersect at $Y$ .
2) $Y$ must be on line $BM$ because $FDCY$ is a parallelogram and its diagonals must intersect at their midpoints.
3) $\angle FYC = \angle ADE$ and $\frac{AD}{FY} = \frac{BD}{BY} = \frac{DE}{CY}$ therefore triangles $FYC$ and $ADE$ are similar and $FC \parallel AE$
4) $\angle MCA = \angle YFC = \angle DAE = \angle DBE$ hence the circumcircle of triangle $BDC$ is tangent to line $MC$ therefore $MC^2 = MD \cdot DB$

Assume $MC^2 = MD \cdot DB$
1) Then $\angle MCD = \angle DBE = \angle DAE$ hence $FC \parallel AE$
2) $\angle MFE = \angle DEA = \angle DBA$ hence the circumcircle of triangle $BDF$ tangent to line $MF$
3) Hence $MF^2 = MD \cdot MB = MC^2 \Rightarrow MF = MC$

Daniel