Extend $AK$ to intersect $BC$ at $X$. Let $X'$ be the foot of the perpendicular from $A$ to $BC$. As $AD$ is the angle bisector we see $DE=DF \, , \, AE=AF$. Using $\triangle DFC \simeq AX'C \, , \, \triangle BED \simeq BX'A$ we see: $$\frac{BE}{BX'}=\frac{ED}{AX'}=\frac{DF}{AX'}=\frac{CF}{CX'} \implies \frac{BE}{FC}=\frac{BX'}{CX'}$$ Now using Ceva we get: $$\frac{BX}{XC}=\frac{AF}{FC} \cdot \frac{BE}{AE}=\frac{BE}{FC}=\frac{BX'}{CX'} \implies X=X'$$So $AK \bot BC$ as desired.

Dear Mathlinkers, http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=203779 Sincerely Jean-Louis

In an acute-angled triangle $\triangle ABC$, $D$ is the point on $BC$ such that $AD$ bisects $\angle BAC$, $E$ and $F$ are the feet of the perpendiculars from $D$ onto $AB$ and $AC$ respectively. The segments $BF$ and $CE$ intersect at $K$. Prove that $AK$ is perpendicular to $BC$.