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

Question

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).

Answer 1

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 $AB$