To prove $AE$ bisects $\angle BAC$

Question

In the figure below, given that $DE\:$ bisects $\angle BDF$ and $FE\:$ bisects $\angle DFC$. Prove that $AE\:$ bisects $\angle BAC$.

My try: I dropped perpendiculars from $E$ to $DB$ and $FC$. But could not figure out.

Answer 1

Drop altitudes from E to AB, DF and AC, with respective feet X, Y and Z.

The right triangles DXE and DYE are congruent; so are FYE and FZE, Thus, EX = EY = EZ and EXA and EZA are congruent, hence the angle bisector AE.

Answer 2

$E$ is an excentre of $\triangle ADF$ -- it is the centre of a circle tangent to all three sides, as $$\operatorname{dist}(E,\overline{AC})=\operatorname{dist}(E,\overline{DF})=\operatorname{dist}(E,\overline{AB}).$$

So automatically we get that $AE$ bisects $\angle DAF$.

Answer 3

Same answer as the others, but with different words:

The circle with center at $E$ that has $AB$ and $DF$ as tangents (which exists because $DE$ is a bisector) is necessarily the same as the circle that has center at $E$ with $AC$ and $DF$ as tangents.

Thus there is a circle centered at $E$ with both $AB$ and $AC$ as tangents, so $AE$ bisects the angle between those lines.