Intercepts Similarity

Let ABD be a triangle with BD the base. Let C be on BD. Construct the line AC. Construct a line parallel to the base. Extend AB to intersect the parallel at E, AC to intersect the parallel at F, and AD to intersect the parallel at G.

I say that BC is to CD as EF is to FG.

Since BD is paralell to EG, and AE, AF, and AG fall upon them, we can say that ABC AEF, ACB AFE, ACD AFG, and ADC AGF.		I.29
Since the angles of triangle ACD are congruent to the angles of AFG, the corresponding sides are proportionate, and AC is to AF as CD is to FG.		VI.4
Again, since the angles of triangle ABC are congruent to the angles of AEF, the corresponding sides are proportionate and AC is to AF as BC is to EF.		VI.4
Since CD is to FG as AC is to AF and AC is to AF as BC is to EF, then CD is to FG as BC is to EF.		V.11
Since CD is to FG as BC is to EF, then CD is to BC as FG is to EF.		V.16
Thus, if two sides of a triangle and a line intersecting the base and apex of that triangle are extended to intersect a line parallel to the base, then the lengths cut in the parallel line are in the same ratio as the lengths cut in the base.
		Q.E.D.

-David