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Van Schooten's Ruler Constructions - Thrid Solution to Problem I

Author(s): 
C. Edward Sandifer

Van Schooten writes

 

And another way.  Assume, as before, AD equals AB, that BD are joined, and that BE equals BA.  And from E through D an indefinitely long straight line is drawn.  In that, if DF is put, equal to DE, and from F through H [sic.  should be A] is drawn FAf:  I say that this bisects the angle BAC.

 

 

 

Branch point

        Fourth (and last ) solution to Problem I

        Problem II.

 

 

C. Edward Sandifer, "Van Schooten's Ruler Constructions - Thrid Solution to Problem I," Loci (August 2010)

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