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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.

figure 1 here

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.

figure 2 here

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

figure 3 here

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).

figure 4 here

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  • $\begingroup$ 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 $\endgroup$
    – Henry
    Jul 9, 2020 at 0:32
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    $\begingroup$ Do you know Ceva's Theorem? If you do, then use it to establish $$\frac{ED}{DA}=\frac{EG}{GB}\,.$$ $\endgroup$
    – Batominovski
    Jul 9, 2020 at 1:41
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    $\begingroup$ 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 $\dfrac{EG}{GB}=\dfrac{EM}{AB}$. $\endgroup$
    – Batominovski
    Jul 9, 2020 at 1:47
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    $\begingroup$ 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. $\endgroup$
    – Mehdi
    Jul 9, 2020 at 5:05
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    $\begingroup$ @Batominovski I suspect you intended to say $N$ was where $\ell$ meets $BD$ rather than $BG$ $\endgroup$
    – Henry
    Jul 9, 2020 at 8:26

1 Answer 1

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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 $AB$
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  • $\begingroup$ A vector based demonstration is interesting. But you might be skipping too much steps on your explanation, that you think are trivial. Especially, how you expressed vector AF as a combination of vector b and d. Unless this was intended to be guidelines. About the shear transformation proof, it looks more intuitive than rigorous to me. I'm unsure, it would be accepted at school. $\endgroup$
    – Mehdi
    Jul 9, 2020 at 12:35
  • $\begingroup$ @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 $t=\frac{k-1}{2k-1}$. $\endgroup$
    – Henry
    Jul 9, 2020 at 13:43
  • $\begingroup$ Ok, I managed to check "quickly" your vector based proof. But there's probably an other typo, I think you meant "G lies on BE and AF". That's not the easiest proof, but it's fine. $\endgroup$
    – Mehdi
    Jul 9, 2020 at 15:48
  • $\begingroup$ @Mehdi - thank you for that - edited $\endgroup$
    – Henry
    Jul 9, 2020 at 15:52

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