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If three points, one on each side of a triangle (extended if necessary) are collinear, then the product of the ratios of division of the sides by the points is -1 if internal ratios are considered positive and external ratios are considered negative.

The proof is is to show that $\dfrac{AG}{GC}\dfrac{CI}{IB}\dfrac{BH}{HA} = -1$

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The following pairs of similiar triangles are observed.

$$\triangle ADG \sim \triangle CFG$$ $$\triangle BEI \sim \triangle CFI$$ $$\triangle BEH \sim \triangle ADH$$

And the proportional sides are

$$\dfrac{AG}{GC} = \dfrac{AD}{CF}$$ $$\dfrac{CI}{IB} = -\dfrac{CF}{BE}$$ $$\dfrac{BH}{HA} = \dfrac{BE}{AD}$$

Then $\dfrac{AG}{GC}\dfrac{CI}{IB}\dfrac{BH}{HA} = \dfrac{AD}{CF} -\dfrac{CF}{BE} \dfrac{BE}{AD} = -1 $

Sorry for the long introduction post, but the proof you see above is Menelaus's theorem shown in my book. I have some concerns over the conventions used.

In particular, I am extremely confused by the equality $$\dfrac{AG}{GC} = \dfrac{AD}{CF}$$

How did they decide to use GC instead of CG or CF instead of FC for that matter? I know that the original equality $\dfrac{AG}{GC}\dfrac{CI}{IB}\dfrac{BH}{HA} = -1$ was derived from going around the triangle counterclockwise.

I even tried to mirror and flip the triangles, that is.

So I should get $\dfrac{AG}{CG} = \dfrac{AD}{CF} = \dfrac{GD}{GF}$

What is wrong with my reasoning?

EDIT: After asking my professor about this, he said it was poor convention and that the book was swapping the magnitudes and the "vector" segments without being explicit. Problem solved...

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