Show that the area of two triangles formed by lines in parallelogram are equal Consider the parallelogram $ABCD$. Let points $E$ and $F$ lie on $AB$ and $AD$, respectively. The line $EF$ meets the extension of $DC$ in the point $H$ and the extension of $BC$ in the point $G$. Show that the $|\Delta AGH| = |\Delta EFC|$ (see figure below).
My first approach was to label the heights of the respective triangles $h_1$ ($\Delta AGH$) and $h_2$ ($\Delta EFC$). If the areas have to be equal, we have:
$$
\begin{equation}
   \frac{|\Delta AGH|}{|\Delta EFC|} = 1 \iff \frac{h_1(|HF| + |EF| + |EG|)}{h_2|EF|} = 1 \iff \frac{h_1}{h_2}(\frac{|HF| + |EG|}{|EF|} + 1) = 1
\end{equation}
$$
But now, I realized that (because of A-A-A for similar triangles), $\Delta HFD\sim\Delta AFE\sim\Delta EGB$, which means that we can rewrite the yet to be proven equation equivalently as:
$$
\begin{equation}
\frac{h_1}{h_2}(\frac{DF}{AF} +\frac{BE}{AE} + 1) = 1
\end{equation}$$
But from here, I couldn't really find any way forward... any help would be deeply appreciated.

 A: Based on your similar triangles observation, let $\def\ta#1{{\left|{\triangle #1}\right|}} a = \ta {AEF}$ as a unit area.
To calculate $\ta{AGH}$,
$$\begin{align*}
\ta{AGH} &= \frac{GH}{EF}\ta{AEF} = \frac{GH}{EF}a
\end{align*}$$
To calculate $\ta{EFC}$, observe that $\triangle AEF \sim \triangle CHG$ by AAA,
$$\begin{align*}
\ta {CHG} &= \left(\frac{GH}{EF}\right)^2\ta{AEF} = \left(\frac{GH}{EF}\right)^2 a\\
\ta{EFC} &= \frac{EF}{GH} \ta{CHG}\\
&= \frac{GH}{EF}a\\
&= \ta{AGH}
\end{align*}$$
A: Let $h_A, h_C$ be the heights of $A, C$ from the line $HG$.
Since $\triangle AEF \sim \triangle CHG$ by AAA (This is the other similar triangle that OP initially missed in the writeup), hence $ \frac{h_A}{h_C} = \frac{EF}{HG}$.
Thus $ |\triangle AHG| = \frac{1}{2} h_a \times HG = \frac{1}{2} h_c \times EG = |\triangle CEF| $.
A: I found the last step from my initial approach as well:
If we want to show that the equation
$
\begin{equation}
\frac{h_1}{h_2}(\frac{|HF| + |EG|}{|EF|} + 1) = 1
\end{equation}
$
holds, it is equivalent as to showing that the value $\frac{|HF| + |EG|}{|EF|}$ is equal to $\frac{h_2}{h_1} - 1$, which can be done by showing that $\frac{h_2}{h_1} - \frac{|HF| + |EG|}{|EF|} = 1$, but since $\Delta HCG\sim\Delta AEF$, we can write $\frac{h_2}{h_1} = \frac{|HF| + |EF| + |EG|}{|EF|}$, which gives:
$
\begin{equation}
\frac{h_2}{h_1} - \frac{|HF| + |EG|}{|EF|} = \frac{|HF| + |EF| + |EG|}{|EF|} - \frac{|HF| + |EG|}{|EF|} = \frac{|EF|}{|EF|} = 1
\end{equation}
$
Which was to be showed
