Proving that $BI$, $AE$ and $CF$ are concurrent? 
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? :)

Consider that: 
$$\frac {AF}{BF}=\frac{FA}{FD}\cdot \frac{FD}{FB}=\cot A\cdot \tan \frac{B}{2}$$
$$\frac {BE}{EC}=\frac{BE}{DE}\cdot \frac{DE}{CE}=\cot \frac{B}{2}\cdot \tan C$$
$$\frac{IC}{IA}=\frac{IC}{IB}\cdot \frac{IB}{IA}=\cot C\cdot \tan A$$
Multiply those equalities side by side, $AE,BF,CI$ are concurrent by Ceva theorem.
