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

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:

• From here, proving that $\overline{EF'} = \overline{DF''}$ would be sufficient as parallelograms are formed. Try using Thale's theorem maybe that will help. Commented Feb 17, 2023 at 1:26
• Yes, that's exactly what I'm trying to show, but I don't see how Thale's theorem applies here, there are no circles. Commented Feb 17, 2023 at 1:29

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.

• Since you’re open to insightful: If that’s to easy then use similar triangles: $$\Delta ADF \sim \Delta ABG\Rightarrow \frac{AF}{AG}=\frac{DF}{BG}$$ $$\Delta AEF \sim \Delta ACG\Rightarrow \frac{AF}{AG}=\frac{EF}{CG}$$ $$\frac{DF}{BG}=\frac{EF}{CG}, DF=EF:BG=CG(\therefore)$$ Commented Feb 17, 2023 at 1:47

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 $$\overline{BC}$$.

• This is a fabulous proof but I am not allowed to use similar triangles ); Commented Feb 17, 2023 at 1:40
• @ClydeKertzer Thanks for letting me know. Since I'm unaware of what the propositions are in Euclid's Elements Book I, II, and III, and I suspect most of the other site members will also be unaware, I suggest you give a summary of what's in them, or perhaps some of the common approaches, such as with similar triangles in my answer, that you're specifically not allowing to be used. Since the doesn't answer your question, I'll delete this post shortly. Commented Feb 17, 2023 at 1:42
• I made an edit to your post, but don't delete yours, it's still an insightful proof. Commented Feb 17, 2023 at 1:44