51st Ukrainian National Mathematical Olympiad, 4th Round

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 $△ATD\sim ETG$, $△GHE\sim CHB$ and $△ASD\sim BSC$. This implies that $$\frac{AT}{TE}=\frac{AD}{GE},\frac{EH}{BH}=\frac{GE}{BC},\frac{BS}{SA}=\frac{BC}{AD}.$$ 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.