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