Circle diameter $BC$ cuts side $AB$ and $AC$ of $\triangle ABC$ respectively $C'$ and $B'$. $E$ and $E'$ are points respectively on $BC$ and the circumcircle of $AB'C'$ such that $EE'$ passes through $A$ and $EE'$ cuts the circle diameter $BC$ at $D$ and $D'$. Prove that $AD \cdot AD' = AE \cdot AE'$.
I tried proving that $AE \cdot AE' = AM^2$ with $AM$ is a tangent of the circle diameter $BC$.
$$\measuredangle ECB'=\measuredangle AC'B'=\measuredangle AE'B',$$ which says that $EE'B'C$ is cyclic.
Id est, $$AD'\cdot AD=AB'\cdot AC=AE'\cdot AE$$ and we are done!