Quadrilateral problem with midpoints of diagonals [duplicate]

For an arbitrary convex quadrialteral $ABCD$:

$$AF=FC, BE=ED, GI=IH$$

Prove that points $F$, $E$ and $I$ are collinear.

I was able to solve the problem by using vectors and I think that it can be also solved analytically but I wanted some more elegant proof, more in the spirit of Euclid.

Quoting Johnson, Advanced Euclidean Geometry:

A projective solution:

Fix points $A$, $B$, $C$ and $G$ and move $D$ on (fixed) line $AG$, so that $H$ moves on (fixed) line $AB$. Point $E$ moves on parallel (middle line) $p$ to $AG$ and point $I$ moves on parallel (middle line) $q$ to $AB$. (Let $p\cap q=\{S\}$, it is easy to see that $S$ halves $BG$!) So $$E\mapsto D \mapsto H\mapsto I$$ is projective transformation from line $p$ to line $q$ since it is composite of $3$ perspective transformation respectively with centers at $B$, $C$ and $G$. Now this transformation is also perspective with respect to some point $T$ since when $E=S$ then $D=G$ and $I=S$, so $S$ maps to it self.

We are left to identify point $T$. And it is not difficult to see that $T=E$.