Van Schooten's Ruler Constructions - Third Solution to Problem II

Finally, van Schooten presented a third solution to Problem II.

Problem II: Given a straight line with endpoints A, B, to bisect it.

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 EF equals CB.  [Here, van Schooten gives five illustrations, showing each of the following five cases:  1) DF = DA, 2) DF < DA, 3) DF > DA but DE < DA, 4) DE = DA, and 5) DE > DA.  We illustrate case #2 only.)]  Make BE, 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.

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