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.
If clarification is needed I'll be glad to answer in the comments.
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$\begingroup$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.
$\endgroup$ 2 $\begingroup$Try for an indirect proof and use the (forwards) theorem on the parallel line in your linked picture.
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