The elements of geometrie of the most auncient philosopher Euclide of Megara. Faithfully (now first) translated into the Englishe toung, by H. Billingsley, citizen of London. Whereunto are annexed certaine scholies, annotations, and inuentions, of the best mathematiciens, both of time past, and in this our age. With a very fruitfull præface made by M. I. Dee, specifying the chiefe mathematicall scie[n]ces, what they are, and wherunto commodious: where, also, are disclosed certaine new secrets mathematicall and mechanicall, vntill these our daies, greatly missed

¶The first Proposition. A perpendicular line drawen from the centre of a circle, to the side of a Pentagon inscribed in the same circle:* 1.1 is the halfe of these two lines taken together, namely, of the side of the hexagon, and of the side of the decagon inscribed in the same circle.

TAke a circle ABC, and inscribe in it the side of a pentagon, which let be BC, and take the centre of the circle, which let be the point D:* 1.2 and frō it draw vnto the side BC a perpendicular line DE: which produce to the point ••. And vnto the line E F put the line EG equall. And draw these right lines CG, CD, and CF. Then I say, that the right line DE (which is drawen from the centre to BC the side of the pentagon) is the halfe of ••he side•• of the decagon and hexagon, taken together.* 1.3 Forasmuch as the line DE is a perpendicular ••nto the line BC: therefore the sections BE and EC shall be equall (by the 3. of the third): and the line EF is common vnto them both; and the angles FEC and FEB, are right an∣gles, by supposition. Wherefore the bases BF and FC are equall (by the 4. of the first). But the line BC is the side of a pentagon, by constructi∣on. Wherefore FC which subtendeth the halfe of the side of the

[illustration]

pentagon, is the side of the decagon inscribed in the circle ABC. But vnto the line FC is, by the 4. of the first, equall the line CG, for they subtend right angles CEG, and CEF, which are con∣tained vnder equall sides. Wherefore also the angles CGE, and CFE, of the triangle CFG, are equall, by the 5. of the first. And forasmuch as the arke FC is subtended of the side of a decagon, the arke CA shall be quadruple to the arke CF: Wherefore also the angle CDA shall be quadruple to the angle CDF (by the last of the six••). And forasmuch as the same angle CDA, which is set at the center, is double to the angle CFA, which is set at the circumference, by the 20. of the third: therefore the angle CFA, or CFD, is double to the angle CD••, namely, the halfe of qua∣druple. But vnto the angle CFD or CFG, is proued equall the angle CGF: Wherefore the outward angle CGF, is double to the angle CDF. Wherefore the angles CDG and DCG, shall be equall. For vnto those two angles the angle CGF is equall, by the 32. of the first. Wherefore the sides GC and GD, are equall, by the 6. of the first. Wherefore also

the line GD is equall to the line FC, which is the side of the decagon. But vnto the right line FE is e∣quall the line EG, by construction. Wherefore the whole line DE is equall to the two lines C•• and FE. Wherefore those lines taken together (namely, the lines DF and FC) shall be double to the line DE. Wherefore the line DE (which is drawen from the centre perpendicularly to the side of the pen∣tagon) shal be the halfe of both these lines taken together, namely, of DF the side of the hexagon, and CF the side of the decagon. For the line DF which is drawen from the centre, is equall to the side of the hexagon, by the Corollary of the 15. of the fourth. Wherefore a perpendicular line drawen from the center of a circle, to the side of a pentagon inscribed in the same circle: is the halfe of these two lines taken together, namely, of the side of the hexagon, and of the side of the decagon inscribed in the same circle: which was required to be proued.

A Corollary.

If a right line drawen perpendicularly from the centre of a circle to the side of a pentagon, be diuided by an extreme and meane proportion: the greater seg∣ment shall be the line which is drawen from the same c••••tre to the side of an equilater triangle inscribed in the same circle. For, that li•••• (drawen to the side of the triangle) is (by the Corollary of the 12. of the thirtenth) the halfe of the line drawen from the centre to the circumference, that is, of the side of the hexagon: Wherefore the residue shall be the halfe of the side of the decagon. For the whole line is the halfe of the two sides, namely, of the side of the hexagon, and of the side of the decagon. But of the side of a decagon and of an hexagon taken together, the greater segment is the side of the hexagon (by the 9. of the thirtenth). Wherefore the greater segment of their halfes shall be the halfe of the hexagon, by the 15. of the fift: which halfe is the perpendicular line draw∣en from the centre to the side of the triangle, by the Corollary of the 12. of the thirtenth.

Notes

• * 1.1

  The first pro∣position after Campane.
• * 1.2

  Construction.
• * 1.3

  Demonstra∣tion.