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

C. Edward Sandifer

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.


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         The other eight problems