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

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 third solution.

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 FA:  I say that this bisects the angle BAC.

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Fourth (and last) solution to Problem I

Problem II