The extension of the median $AM$ of the triangle $ABC$ intersects its circumcircle at $D$. The circumcircle of triangle $CMD$ intersects the line $AC$ at $C$ and $E$.The circumcircle of triangle $AME$ intersects the line $AB$ at $A$ and $F$. Prove that $CF$ is the altitude of triangle $ABC$.
We have $\angle ABC=\angle ADC= \angle AEM =\angle BFM$ so $\triangle MFB$ is isosceles $\Longleftrightarrow MB=MF=MC$, so $M$ is the cicrumcenter of $\triangle CBF $ wich implies perpendicularity $.\blacksquare$