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? :)

$$\frac{CE}{BE}\cdot\frac{BF}{FA}\cdot\frac{AI}{IC}= \frac{CD\cos C}{BD\cos B/2}\cdot
\frac{BD\cos B/2}{AD\cos A}\cdot\frac{AB\cos A}{BC\cos C}=\frac{CD}{AD}\cdot\frac{AB}{BC}=\frac{BC}{AB}\cdot\frac{AB}{BC}=1$$
Applying mutual Ceva's theorem, statement is demonstrated.
