Show that 3 points are aligned Let $ABCD$ be a quadrilateral in $\mathbb{A}^2$ and call the intersection of $AB$ and $CD$ : $E$ and then the one of $AD$ and $BC$ : $F$. Let $G$ and $H$ two points such that $ABGD$ is a parallelogram and $AEHF$ too.
I want to show that $C,G,H$ are aligned.
So I set the affine coordinate system $(A, E, F)$. Then I worked out $GH$ using the parametrization $\forall t \in \mathbb{R}, ~tG+(1-t)H$. One easy thing I found is $G=A+\overrightarrow{AB}+\overrightarrow{AD}=D+B-A$ so $G$ only depends on points with at least one zero coordinate. Here my goal is to show that $C$ is on $GH$ by finding coordinates of $C$ that depends only of $B$ and $D$ but I don't see how to do that.
 A: Let the intersection of $DG$ with $CF$ be $F'$, intersection of $BG$ with $CE$ be $E'$.
Then by similar triangle, we have $\frac {CF'}{CD}=\frac {CB}{CE}$, then $\frac {CF'}{CF}=\frac {CB\cdot CD}{CF\cdot CE}$.
Similarly, $\frac {CE'}{CE}=\frac {CB\cdot CD}{CF\cdot CE}$, then $\frac {CF'}{CF}=\frac {CE'}{CE}$, showing that the quadrilateral $CE'GF'$ and $CEHF$ are similar.
A: We can use oblique coordinates wrt axes AB and AC (as you implicitly began to do):

Solving linear system
$$\begin{cases}\text{Line BF:} \ &\dfrac{x}{b}+\dfrac{y}{f}&=&1\\ \text{Line DE:} \ &\dfrac{x}{e}+\dfrac{y}{d}&=&1\end{cases}$$
gives the coordinates of their intersection point:
$$C(u,v)=\left(\dfrac{be(f-d)}{ef-bd};\dfrac{df(e-b)}{ef-bd}\right).$$
It remains to use the classical criteria for alignment of 3 points $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ which is
$$\begin{vmatrix}x_1& x_2 &x_3\\
   y_1& y_2 &y_3\\
   1& 1 &1\end{vmatrix}=0$$
Here it becomes
$$\begin{vmatrix}b& e &(be(f-d))\\
   d &f& (df(e-b))\\
   1& 1 &(ef-bd)\end{vmatrix} \ \ \text{which is indeed} \ \ 0.$$
