Show that the area of two triangles formed by lines in parallelogram are equal

Question

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.

Answer 1

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,

$$\begin{align*} \ta {CHG} &= \left(\frac{GH}{EF}\right)^2\ta{AEF} = \left(\frac{GH}{EF}\right)^2 a\\ \ta{EFC} &= \frac{EF}{GH} \ta{CHG}\\ &= \frac{GH}{EF}a\\ &= \ta{AGH} \end{align*}$$

Answer 2

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| $.

Answer 3

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