|Book IV Proposition 16|
BC be bisected at E;
each of the circumferences BE, EC is a fifteenth of the circle ABCD.
If therefore we
join BE, EC
fit into the circle ABCD straight lines equal to them and in contiguity, a fifteen-angled figure which is both equilateral and equiangular will have been inscribed in it.
Being what it was required to do.
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