Show that in figure bisector of angle intersect at perpendicular The following was asked by a high school student which I could not answer. Please help
In the figure below, show that the bisector of $\angle AEB$ and $\angle AFD$ intersect at perpendicular 
 A: Here is a better figure:

Here $ABF, DCF, ADE, BCE$ are collinear, and $PE, PF$ bisects $\angle AEB, \angle AFD$ respectively.
We have:
\begin{align}180^\circ &= \angle BAE + \angle AEB + \angle EBA \\&= \angle BAE + \angle AEB + \angle AFC + \angle BCF\\&=\angle BCF + 2\angle PEB + 2\angle PFC + \angle BCF\end{align}
This shows that $\angle BCF + \angle PEB + \angle PFC = 90^\circ$.
Now consider the quadrilateral $PFCE$ and you will finish the proof:
\begin{align}360^\circ &= \angle PFC + \text{reflex}\angle FCE + \angle CEP + \angle EPF\\
&=\angle PFC + (180^\circ + \angle BCF) + \angle BEP + \angle EPF\\
&=180^\circ + 90^\circ + \angle EPF
\end{align}
giving $\angle EPF = 90^\circ$.
A: 
We claim that $\triangle FHP \cong \triangle FGP$. So $\angle FPG = \angle FPH$. So $\angle FPE = 90^0$.
Now how are they congruent?
$\angle FHP = \angle FAE + \angle AEH$
Also, $\angle FGP = \angle DCE + \angle CEP = \angle FAE + \angle AEH$
So, $\angle FHP = \angle FGP$ and we know $\angle PFH = \angle PFG$.
