This questions is related to the question before. I was mistaken. Because I thought ED and AB are parallel, but they're not. So this question made me really confused and curious about it.
Given a Triangle ABC. AD, BE, and CF are altitudes. Prove that AD is the bisector of FDE angle.
First, let's prove that triangle $CED$ is similar to $CBA$. Since $\angle AEB = 90^{\circ} = \angle ADB$, quadrilateral $ABDE$ is inscribed into a circle. Therefore, $\angle ABD = 180^{\circ} - \angle AED = \angle CED$. By analogy, $\angle BAE = \angle CDE$ what completes the proof. The same goes for similarity of $BDF$ and $BAC$. We want to prove that $\angle FDA = \angle ADE \Leftrightarrow 90^{\circ} - \angle FDA = 90^{\circ} - \angle ADE \Leftrightarrow \angle BDF = \angle CDE$. Since both of them are equal to $\angle BAC$ (due to lemma), proof is complete.
Name the intersection point as $X$
As $DXEC$ is a cyclic quadrilateral(Why?) $$\angle XDE=\angle XCE$$
As $DXFB$ is a cyclic quadrilateral(Why?) $$\angle XDF=\angle XBE$$
As $BFEC$ is a cyclic quadrilateral(Why?) $$\angle FBE=\angle FCE$$
Bingo!
