In $\triangle ABC$, $DE$ parallel to $BC$, $F$ midpoint of $DE$, $AF$ meets $BC$ at $G$. Prove $G$ is the midpoint of $BC$

Question

This question comes from Hartshorne's Geometry: Euclid and Beyond, exercise $3.7$.

$3.7$ Given $\triangle ABC$, let $\overline{DE}$ be a line parallel to the base $\overline{BC}$, let $F$ be the midpoint of $\overline{DE}$, and let $\overline{AF}$ meet $\overline{BC}$ at $G$. Prove that $G$ is the midpoint of $\overline{BC}$.

There is a hint in the book that says

Hint: Draw some extra lines to make parallelograms, and use (I.$43$).)

In case anyone is wondering I.$43$ says

In any parallelogram the complements of the parallelograms about the diameter equal one another.

There are a few proofs of this online, but I am trying to prove it using only the propositions in Euclid's Elements Book I, II, and III. This is what I currently have of my proof but I can't seem to finish it off. I feel like this is the right direction. The goal of the proof is to make parallelograms and leverage their properties to prove the exercise.

1. Given that $\overline{AG}$ bisects $\overline{DE}$ at point $F$, $\overline{DF} = \overline{FE}$.
2. The alt-int-angles and ex-int-angles of $\overline{AG}$ are equal.
3. Move $\overline{FG}$ to $B$ and $C$ such that the segments are parallel to $\overline{AG}$.

Here is what I have written up in tikz so far:

Answer 1

Start by constructing the tan lines parallel to AG, such that the yellow highlighted triangles are congruent. Green highlights are also congruent.

Then observe that tan segments are not only parallel and congruent, but also halved, indicating F as intersection of diagonals in magenta parallelogram.

Therefore $FF_1=FF_2\Leftrightarrow FD-DF1=FE-EF_2, FD=FE:DF_1=EF_2$

From here is straightforward.

Answer 2

Due to $DE \parallel BC$, we have $\measuredangle ADF = \measuredangle ABG$ and $\measuredangle AFD = \measuredangle AGB$, so $\triangle ADF \sim \triangle ABG$, and similarly $\triangle AEF \sim \triangle ACG$. Thus, we have

$$\frac{|DF|}{|AF|} = \frac{|BG|}{|AG|} \; \; \to \; \; |BG| = \frac{|DF||AG|}{|AF|} \tag{1}\label{eq1A}$$

$$\frac{|EF|}{|AF|} = \frac{|CG|}{|AG|} \; \; \to \; \; |CG| = \frac{|EF||AG|}{|AF|} \tag{2}\label{eq2A}$$

Since $|DF| = |EF|$, we get from \eqref{eq1A} and \eqref{eq2A} that

$$|BG| = |CG| \tag{3}\label{eq3A}$$

i.e., $G$ is the midpoint of $\overline{BC}$.