Proof regarding a parallelogram and a given line segment parallel to its side

Question

In parallelogram $ABCD$ there is given a segment $\overline{EF}$ s.t. $\overline{EF}\parallel\overline{BC}.\;$ If $G$ is the intersection point of $BE$ and $CF$ and $H$ is the intersection point of $AE$ and $DF$, prove that $GH\parallel AB$.

My attempt:

Let $J$ be the intersection point of $AE$ and $BC$ and $I$ the intersection point of $DF$ and $BC$.

Then, $\Delta AHD{\sim}\Delta HIJ{\sim}\Delta FEH$.

Analogously, let $K$ be the intersection point of $BE$ and $AD$ and $L$ be the intersection point of $CF$ and $AD$.

Then, $\Delta BCD{\sim}\Delta GKL{\sim}\Delta EFG$.

However, I wasn't sure how to use those similarities.

I also considered the following:

Let $M$ be the intersection point of $AE$ and $CF$.

Then $\Delta AML{\sim}\Delta EMF{\sim}\Delta CJM$

May I ask for advice on solving this task? Thank you in advance

Answer 1

One way to prove this statement is with a use of homothety.

Observe a homothety $\mathcal{H}_H$ at $H$ that takes $A$ to $E$ and $D$ to $F$ and a homothety $\mathcal{H}_G$ at $G$ that takes $E$ to $B$ and $F$ to $C$.

Then their composition $\mathcal{T}:=\mathcal{H}_G\circ \mathcal{H}_H$ takes $A$ to $B$ and $D$ to $C$ which means that the center of new homothety $\mathcal{T}$ is on $AB$ and $CD$ which are parallel. But this can only be point in infinty which is also on a line $GH$ which means $AB||GH$.

Note:

This means that $\mathcal{T}$ is actually a translation for a vector $\overrightarrow{AB}$.

Answer 2

Second solution is with a use of Desargues theorem:

Lines $AD$, $BC$ and $EF$ are perspective with respect to a point in infinity (which is on the line $AD$). So the triangles $ABE$ and $CDF$ are perspective to a line $EF$, which means that intersection point $X$ of $AB$ and $CD$ is on $EF$. But $X$ is infinity point on $AB$ so $AB||EF$.