# Prove that $AD \cdot AD' = AE \cdot AE'$.

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!