# Proof for "equivalence" between mid-points and parallels. Construction using a ruler.

Let's say we have a segment with a mid-point. By exploiting that mid-point, and using only a ruler, I can draw a parallel from any point to that segment. But I still have to demonstrate that it is a parallel.
So in a first time I'll show you how I draw this parallel, and then I'll ask for a proof. In the case that's wrong, provide a proof too.

I can't embed pics, see links.

At start (see fig 1), we have a segment AB. The mid-point is C (i.e length or AC and CB are same). We have an arbitrary point D. We want to draw a parallel line to segment AB, that go through point D, using only a ruler.

So we draw the half-line AD with origin A and that go through point D (see fig 2). Let's take an arbitrary point E, on that half-line, but not on AD segment. Then, let's draw segment BE.

We draw then segment CE and BD. (see fig 3). This creates a new intersection point, that we call F.

Now let's draw half-line AF with origin A and that go through point F (see fig 4). There's a new intersection point between half-line AF and segment BE that we call G. Finally let's draw line DG that go through point D and G. Please, provide proof that line DG is a parallel to segment AB (otherwise demonstrate contrary).

• What kind of proof do you want. For example, if you do a shear transformation to your figure 4 based on $AB$ which moves $E$ to be on the perpendicular bisector of $AB$ then it is true by symmetry. Now undo the shear transformation, and it is still true Jul 9, 2020 at 0:32
• Do you know Ceva's Theorem? If you do, then use it to establish $$\frac{ED}{DA}=\frac{EG}{GB}\,.$$ Jul 9, 2020 at 1:41
• If not, let $AG$ meet the line $\ell$ parallel to $AB$ passing through $E$ at $M$, and let $BG$ meet $\ell$ at $N$. Show that $EM=EN$, $\dfrac{ED}{DA}=\dfrac{EN}{AB}$, and $\dfrac{EG}{GB}=\dfrac{EM}{AB}$. Jul 9, 2020 at 1:47
• Thanks for suggesting Ceva's Theorem. (I've just learned it). So when combined with Reciprocal of Intercept Theorem, the demonstration is straightforward. However, I might have misunderstood second comment: when I follow the steps, I see that the points N and E are same. Jul 9, 2020 at 5:05
• @Batominovski I suspect you intended to say $N$ was where $\ell$ meets $BD$ rather than $BG$ Jul 9, 2020 at 8:26

There are many possible proofs: Batominovski gave a couple in the comments based on classical geometry and I gave another saying "if you do a shear transformation to your figure 4 based on $$AB$$ which moves $$E$$ to be on the perpendicular bisector of $$AB$$ then it is true by symmetry. Now undo the shear transformation, and it is still true."
Here is another, based on vectors from $$A$$:
• If $$AB$$ is $$\vec{b}$$ then $$AC$$ is $$\frac12\vec{b}$$
• If $$AD$$ is $$\vec{d}$$ and $$AE$$ is $$k\vec{d}$$ for some $$k$$, then $$AF$$ is $$\frac{k-1}{2k-1}\vec{b} + \frac{k}{2k-1}\vec{d}$$ as $$F$$ lies on $$BD$$ and $$CE$$
• So $$AG$$ is $$\frac{k-1}{k}\vec{b} + \vec{d}$$ as $$G$$ lies on $$BE$$ and $$AF$$
• and thus $$DG$$ must be parallel to $$AB$$
• @Mehdi The skipped steps are not trivial but they are simple. For example $AF$ must be $t\vec{b}+(1-t)\vec{d}$ for some $t$ as $F$ lies on $BD$ and also $s\frac12\vec{b}+(1-s)k\vec{d}$ for some $s$ as $F$ lies on $CE$. Solving this by matching coefficients gives two simultaneous equations leading to $s=2t$ and thus $t=\frac{k-1}{2k-1}$. Jul 9, 2020 at 13:43