Another way: Assume, as before, C is a point away from AB, and from B through C is drawn an indefinitely long line, and find on it CD equal to CB, and join it to A to form AD. Then in DA, assume DE equals DC and DF equals CB. [Here, van Schooten gives five illustrations, showing each of the following five cases: 1) DF = DA, 2) DF < DA, 3) DV > DA but DE < DA, 4) DE = DA, and 5) DE > DA. We illustrate case # 2)] Make VE, FC intersecting at G, then DGH cutting FB in H.
If now F falls on the point A, then the lines FB and AB coincide, and then H bisects it.
But if the point F falls beyond or within A, I say that if the line is drawn through the points C and H, it will bisect the line AB at I.