Prove converse Thales theorem, proportional sides imply parallel lines

I'm going through John Stillwell's Four Pillar's of Geometry and trying to follow the book's structure when doing the exercises. Generally, a 'pillar' is divided into two chapters; the first chapter states useful results and motivation while the second chapter brings the machinery to do the proofs.

Thales' Theorem: Suppose that the line EF is parallel to BC. Then AE/EB=AF/FC.

Question

Prove Converse Thales' Theorem: Suppose that EG is not parallel to BC. Then AE/EB not equal to AG/GC. Equivalently, AE/EB=AG/GC is sufficient for EG parallel BC.

The issue is that I don't think I'm 'allowed' to use much. The ideas of similar triangles, SAS, etc. have not been developed. I know that without having the book in front of you it's hard to judge what can be used but essentially I'm looking for the dirt simplest proof. From what I can see the only known facts are Thales theorem and that was only stated.

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In view of the fact that I do not have access to a copy of the book, probably I should say nothing. But suppose the line segment "looks" like the one drawn (but no $F$ yet). Draw the line $EF$ parallel to $BC$. (Is it already "known" that there is such a line? Formally we need a version of Euclid's Axiom of Parallels.) If $F \ne G$, it is "obvious" (??) that $AG/GC \ne AF/FC$. –  André Nicolas May 19 '11 at 17:22
@user6312 My comment to the answer might answer this more. As it happens Stillwell does take the existence and uniqueness of parallel lines as a given. Also, F not G works like you have it so far as I can tell. It seems like it comes down to realizing that you are cutting the line at a different point and then it becomes obvious that the converse is true. –  Tony May 19 '11 at 19:06

Proof of the converse of "Thales Theorem":

First, let the line $d$ intersect the sides $AB$ and $AC$ of $\triangle ABC$ at distinct points E and G, respectively, such that $\displaystyle \frac{AE}{EB} = \displaystyle \frac{AG}{GC}.$ We now need to prove that $EG \parallel BC$.

Assume $EG \nparallel BC$. Then there must be another line intersecting point $E$ of side $AB$ as well as some point, say $F$, of side $AC$ that is parallel to $BC$. So, let $EF \parallel BC$.

By Thales Theorem, since $EF \parallel BC$, it follows that: $$\frac {AE}{EB} = \frac{AF}{FC}\quad\quad (1)$$ But we are given $$\frac{AE}{EB} = \frac{AG}{GC}\quad\quad (2)$$ Hence, from (1) and (2), it must follow that $$\frac{AF}{FC} = \frac{AG}{GC}\quad\quad (3)$$ Adding "1" to both sides of equation (3) gives us: $$\frac {AF}{FC} + \frac{FC}{FC} = \frac {AG}{GC} + \frac{GC}{GC}$$ which simplifies to $\displaystyle \frac{AF+FC}{FC} = \frac{AG+GC}{GC} \implies \frac{AC}{FC} = \frac{AC}{GC} \implies FC = GC$

But $FC = GC$ is only possible when points $F$ and $G$ coincide with one another, i.e. if $EF$ is the line $d = EG$ itself.

But $EF\parallel BC$, and hence it cannot be the case that $EG = EF \nparallel BC$.

Therefore, if a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side: i.e. the converse of Thales Theorem has been established.

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Note, the first part of your question isn't correct, I was only trying to state the problem not prove it! I'll be more clear. This is a great answer and fills in what I didn't in my comment on user9325's. Thanks. –  Tony May 19 '11 at 22:17
I got it. I thought that was your reasoning for proving the theorem! I'll edit my answer. I hope I didn't get lines and points mixed up. My first "penciled" proof had the triangle oriented differently, so I had to correct the lines ;-) Also, I made use of a "betweenness" axiom: assumed by Euclid, and not made explicit until Pasch, then Hilbert: If three points lie on the same line (e.g. A, E, B), then E lies between A and B if and only if AE + EB = AB (it's also a degenerate form of the triangle equality, but you don't need to know that!). –  amWhy May 19 '11 at 22:41
@Tony: This is absolutely correct. But it is possible to formulate the proof without a contradiction: Just draw the parallel, look at both ratios and conclude that, in fact, $F=G$. –  Phira May 19 '11 at 20:56