Let the side of the inscribed quinquangle be ae: of the sexangle, ei: Of the decangle ao. I say, the side ae, doth in power countervaile the rest. For let there be two perpēdiculars: The first io, the second iu, cutting the sides of the quinquangle and decangle into halves: And the meeting of the second perpendicular with the side of the quinquangle let it be y. The syllogisme of the demonstration is this: The oblongs of the side of the quinquangle, and the segments of the same, are equall to the quadrates of the other sides. But the quadrate of the same whole side, is equall to the oblongs of the whole, and the segments, by the [3 e, xiij]. Therefore it is equall to the quadrates of the other sides.
Let the proportion of this syllogisme be demonstrated: For this part onely remaineth doubtfull. Therefore two triangles, aei, and yei, are equiangles, having one common angle at e: And also two equall ones aei, and eiy, the halfes, to wit, of the same eis: Because that is, by the [17 e, vj]: one of the two equalls, unto the which eis, the out angle, is equall, by the [15 e. vj]. And this doth insist upon a halfe periphery. For the halfe periphery als, is equall to the halfe periphery ars: and also al, is equall to ar. Therefore the remnant ls, is equall to the remnant rs: And the whole rl, is the double of the same rs: And therefore er, is the double of eo: And rs, the double of ou. For the bisegments are manifest by the [10 e, xv]. and the [11 e, xvj]. Therefore the periphery ers, is the double of the periphery eou: And therefore the angle eiu, is the halfe of the angle eis, by the [7 e, xvj]. Therefore two angles of two triangles are equall: Wherefore the remainder, by the [4 e vij], is equall to the remainder. Wherefore by the [12 e, vij], as the side ae, is to ei: so is ei, to ey. Therefore by the [8 e xij], the oblong of the extreames is equall to the quadrate of the meane.
Now let oy, be knit together with a straight: Here againe the two triangles aoe, and aoy, are equiangles, having one common angle at a: And aoy, and oea, therefore also equall: Because both are equall to the angle at a: That by the [17 e, vj]: This by the [2 e, vij]: Because the perpendicular halfing the side of the decangle, doth make two triangles, equicrurall, and equall by the right angle of their shankes: And therefore they are equiangles. Therefore as ea, is to ao: so is ea, to ay. Wherefore by the [8 e, xij]. the oblong of the two extremes is equall to the quadrate of the meane: And the proposition of the syllogisme, which was to be demonstrated. The converse from hence as manifest Euclide doth use at the 16 p xiij.
16. If a triangle and a quinquangle be inscribed into the same Circle at the same point, the right line inscribed betweene the bases of the both opposite to the said
point, shall be the side of the inscribed quindecangle. 16. p. iiij.
For the side of the equilaterall triangle doth subtend 1/3 of the whole pheriphery. And two sides of the ordinate quinquangle doe subtend 2/5 of the same. Now 2/5 - 1/3 is 1/15: Therefore the space betweene the triangle, and the quinquangle shall be the 1/15 of the whole periphery.
Therefore
17. If a quinquangle and a sexangle be inscribed into the same circle at the same point, the periphery intercepted beweene both their sides, shall be the thirtieth part of the whole periphery.