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. enter image description here

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


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.

  • $\begingroup$ 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 $AG/GC \ne AF/FC$. $\endgroup$ May 19, 2011 at 17:22
  • $\begingroup$ @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. $\endgroup$
    – ttt
    May 19, 2011 at 19:06

2 Answers 2


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.

  • $\begingroup$ 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. $\endgroup$
    – ttt
    May 19, 2011 at 22:17
  • $\begingroup$ 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!). $\endgroup$
    – amWhy
    May 19, 2011 at 22:41

Try for an indirect proof and use the (forwards) theorem on the parallel line in your linked picture.

  • $\begingroup$ Am I understanding your advice correctly? Suppose EG is not parallel to BC. Draw the parallel to BC through E, call this EF. Both EG and EF share E, however only EF is parallel to BC so F is not equal to G. By Thales, AE/EB=AF/FC. Since F isn't equal to G AE/EB not equal AG/GC. (Indirectly, I guess you are talking about assuming that AE/EB=AG/GC and saying this is a contradiction since F not G.) $\endgroup$
    – ttt
    May 19, 2011 at 19:02
  • $\begingroup$ @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, $F=G$. $\endgroup$
    – Phira
    May 19, 2011 at 20:56

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