how to prove that the segment $IF=HF+GF$ 
$AE$ and $CD$ are the angle bisectors of $\triangle ABC$. $F$ is an arbitrary point on line $DE$. Prove that $GF+HF=IF$.

I noticed $3$ cyclic quadrilaterals. Any ideas.
Here is the picture
 A: Consider Trilinear Coordinates (https://en.wikipedia.org/wiki/Trilinear_coordinates) first in the case where $F$ is inside triangle $ABC$.
$D$ and $E$, being feet of angle bissectors, have resp. trilinear coord. $(1,1,0)$ and $(0,1,1)$. Therefore, the trilinear equation of straight line $DE$ is:
$$\begin{vmatrix}1&0&x\\1&1&y\\0&1&z\end{vmatrix}=0 \ \ \iff \ \ x-y+z=0\tag{0}$$
Interpreting $(x=FG,y=FH,z=FI)$, we get:
$$FG+FI-FH=0\tag{1}$$
(which is not the given relationship !)
Now, if $F$ is not inside triangle $ABC$, here are the other cases:

*

*In the case depicted in the given figure ($F$ "just outside" $[DE]$ on the side of $E$), only one of the trilinear coordinates, $FG$, undergoes a sign change ; therefore (1) becomes:

$$\color{red}{-}FG+FI-FH=0\tag{2}$$
which amounts to the given relationship, this time !
If, in the case of the given figure, $F$ is far away, a second sign change occurs, now for signed distance $FH$, transforming (2) into  :
$$-FG+FI\color{red}{+}FH=0\tag{3}$$
which is a third formula.

*

*if, on the contrary, $F$ is outside of line segment $[D,E]$ but on the side of $D$, we have to change $FI$ into its opposite in (1), giving back relationship (3).

Remark about relationship (0): we have obtained it by working up to a multiplicative constant ; this is unimportant because we deal with relationships having a zero in their right-hand side.
