Let $ABC$ be a triangle, and $BD$ be the angle bisector of $\angle B$. Let $DF$ and $DE$ be altitudes of $\triangle ADB$ and $\triangle CDB$ respectively, and $BI$ is an altitude of $\triangle ABC$. Prove that $AE$, $CF$ and $BI$ are concurrent.
I saw that $BEDF$ is a kite as well as a cyclic quadrilateral, and even $I$ would lie on that circle. My approach was to assume $H$ as the intersection of $AE$ and $CF$, and prove that $BH\perp AC$, but it didn't work. Can anyone help? :)