Olympiad math problem - Show that a pair of lines is parallel 
Let $ABCD$ be a parallelogram. Draw a circunference that passes through $A$. It intersects $AB$,$AD,$ and $AC$ (for the second time), at points $E,F,$ and $G$, respectively. Lines $EG$ and $FG$ intersect $DC$ and $CB$ at $H$ and $I$, respectively.
Show that $EF$ is parallel to $HI$.

 A: $AEGF$ is cyclic, and hence $\angle GAF=\angle GEF$.  
Next $\angle EGF=\angle HGI$, and hence $GHCI$ is also cyclic, and hence
$$
\angle GEF=\angle GAF=\angle BCA=\angle HIG
$$
and thus $FE\parallel HI$.
Note. The conclusion is still true even in the case in which $A$ is the centre of the circle. Indeed,
triangles $AGE$ and $CGH$ are similar, and hence both isosceles, and hence $CH=CG$. Also, triangles $AGF$ and $CGI$ are similar, and hence both isosceles, and hence $CI=CG$. Thus, 
$C$ is the centre of the circle which is defined by the point $H$, $G$ and $I$. Thus 
$$
2\angle FEH=\angle DAC = \angle GCI=2\angle EHI
$$
and hence $FE\parallel HI$.
A: This problem has nothing to do with a circle. $E,F,G$ can be arbitrary, they don't have to be on the same circle. with $A$. 

Remember that a homothety takes line to a line parallel to it. 
Observe a homothety $\mathcal{H}$ with center at $G$ that takes $F$ to $I$. Then it takes line $AD$ to a line $BC$ since they are parallel. Since it take line $AC$ to it self it follows that it takes $A$ to $C$ and thus line $AB$ to a line $CD$  since they are also parallel. Since it takes line $EH$ to it self we see that it takes $E$ to $H$. So it takes line $EF$ to a line $HI$ which means that $EF||HI$ and we are done.
