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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 $\dfrac{AG}{GC}\dfrac{CI}{IB}\dfrac{BH}{HA} = -1$

enter image description here

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...

enter image description here

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