T H E E L E M E N T S |
Book IV Proposition 16 | |
Let BC be bisected at E; therefore each of the circumferences BE, EC is a fifteenth of the circle ABCD. If therefore we join BE, EC and 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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