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let there be inscribed a side AC of the equilateral triangle inscribed in it,
and
a side AB of an equilateral pentagon.
Therefore, of the equal segments of which there are fifteen in the circle ABCD,
there will be five in the circumference ABC which is one-third of the circle,
and
there will be three in the circumference AB which is one-fifth of the circle;
therefore
in the remainder BC there will be two of the equal segments.
