Prove that $\angle{BED} = \angle{CFD} = 30^{\circ}$ 
Let $D, E$ and $F$ be on $\overline{BC}, \overline{AC},$ and $\overline{AB},$ respectively, such that $\overline{AD}, \overline{BE},$ and $\overline{CF}$ are the angle bisectors of $\triangle {ABC}$. If $\angle{EDF}=90^{\circ}$, prove that $\angle{BED} = \angle{CFD} = 30^{\circ}$.

I have a few conjectures on how prove this. Firstly, it seems that $DE$ is an angle bisector of $\angle{CDA}$ and similarly $DF$ is an angle bisector of $\angle{BDA}$. Also it seems that $\overline{EF} \parallel \overline{BC}$, so that might help, but I don't see how to show $\angle{BED} = \angle{CFD} = 30^{\circ}$.

 A: Note that the quadruple $C,K,G,F$ is harmonic. This can be easily seen via Menelaos and Ceva theorem. 
Since $\angle GDF = \frac \pi 2$ and $C,K,G,F$ is harmonic, it follows that $DG, DF$ are bisectors of the angle between lines $CB, AD$. To see that, let $C'$ be a point on $CK$ such that $DK, DC'$ are symmetric with respect to $DG$. Then $DF, DG$ are angle bisectors of the angle between $DK, DC'$. Using angle bisector theorem we find that $\frac{KG}{GC'} = \frac{KD}{DC'} = \frac{KF}{FC'}$ and so $C',K,G,F$ is harmonic. Therefore $C'=C$.
Since $F$ lies on the internal bisector of angle $ECD$ and on the external bisector of angle $CDE$, it follows that $F$ is $C$-excenter of triangle $CAD$. This means that $\angle FAD = \frac{\pi - \angle DAC}2 = \frac{\pi - \angle FAD}2$. Therefore $\angle FAD = \frac \pi 3= \angle DAC$. Now, angle chasing gives
$$\angle DFC = \pi - \angle FDC - \angle ADF = \pi - \frac{\angle ACD}2 - \angle CDA - \frac{\pi - \angle CDA}2 = \frac \pi 2 - \frac{\angle ACD + \angle CDA}2 = \frac \pi 2 - \frac{\pi - \angle DAC}2 = \frac \pi 6.$$
