We repeat Problem I, trisecting a given angle.
Problem I. Given a rectilinear angle BAC, to cut it into two equal pieces.
Van Schooten wrote a fourth solution.
Again another way: Assume in AB the given point G, and from A is located in AC a line AF equal to AB; CA is produced to E so that EA is either larger or smaller than AB. Then produce BA from the point A and place a straight line AD equal to AE; and make through the points E, B and D, F straight lines EBG and DFG, which intersect at G. This done, if the straight line is drawn through AG, it will bisect that angle BAC.
![](/sites/default/files/images/cms_upload/Sandifer060-115194.jpg)
Van Schooten gives two diagrams, depending on whether EA is larger or smaller than AB. We give here the illustration taking EA larger than AB.
Next:
Problem II
C. Edward Sandifer, "Van Schooten's Ruler Constructions - Fourth Solution to Problem I," Convergence (August 2010)