Without loss of generality, assume that $BC < AD$. Let $S$ be the point of any intersection of the lines $AB$ and $CD$, and $T$ the point of intersection of the lines $AF$ and $DG$ (fig. 34).

First we prove that the points $S$, $H$, and $T$ are collinear. To this end, we use Menelaus' theorem for the triangle $ABE$ and three points $S$, $H$, $T$, that lie on the lines that contain its sides: the points $S$, $H$, $T$ will be collinear if and only if
$$\frac{AT}{TE} \cdot \frac{EH}{HB} \cdot \frac{BS}{SA} = 1.$$
Because $EG \parallel AD$, $GE \parallel BC$ and $AD \parallel BC$, we have that $\triangle ATD \sim ETG$, $\triangle GHE \sim CHB$ and $\triangle ASD \sim BSC$. This implies that
$$\frac{AT}{TE} = \frac{AD}{GE}, \quad \frac{EH}{BH} = \frac{GE}{BC}, \quad \frac{BS}{SA} = \frac{BC}{AD}.$$
So,
$$\frac{AT}{TE} \cdot \frac{EH}{HB} \cdot \frac{BS}{SA} = \frac{AD}{GE} \cdot \frac{GE}{BC} \cdot \frac{BC}{AD} = 1,$$
which proves that the points $S$, $H$, $T$ are collinear.

Next, consider the triangles $AFI$ and $DGC$. Because $IF \cap CG = H$, $FA \cap GD = T$, $AI \cap CD = S$, and these points are collinear, the Desargues' theorem implies that the lines $CI$, $FG$ and $AD$ are concurrent.