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 10. Probleme. The 10. Proposition. To make a triangle of two equall sides called Isosceles, which shall haue eyther of the angles at the base double to the o∣ther angle.

TAke a right line at all aduentures which let be AB, & (by y^e 11. of the second) let it be so deuided in y^e pointe C,* 1.1 y^t the rectangle figure comprehended vnder the lines AB and BC be equall vnto y^e square which is made of the line AC. And making the centre the point A, & the space AB, describe

[illustration]

(by the 3. peticion) a circle BDE, and (by the 1. of the fourth) into the circle BDE apply a right line BD equall to the right lyne AC which is not greater then the diameter of the circle BDE. And draw lines from A to D and from D to C. And (by the 5. of y^e fourth) about the triangle ACD describe a circle ACDF.* 1.2 And forasmuch as the rect∣angle figure contained vn∣der the lines AB and BC is equall to the square which is made of the line AC: (For that is by supposition) But the line AC is equall vnto the line BD. Wherfore that which is contayned vnder the lines AB and BC is equall to the square which is made of the line BD. And forasmuch as without the circle ACDF is taken a poynt B, and from B vnto the circle ACDF are drawen two right lines BCA, and BD, in such sort that the one of them cutteth the circle, and the other endeth at the circumfe∣rence, and that which is contained vnder the lines AB and BC is equall to the square which is made of the line BD, therfore (by the 17. of the third) the line BD toucheth the circle ACDF. And forasmuch as the line BD toucheth in the point D, and from D where the touche is, is drawen a right line DC, there∣fore (by the 32. of the same) the angle BDC is equall vnto the angle DAC, which is in y^e alternate segment of y^e circle. And forasmuch as y^e angle BDC is e∣qual vnto y^e angle DAC, put the angle CDA common vnto thē both. Wherfore y^e whole angle BDA is equal to these two angles CDA, & DAC. But vnto y^e angles CDA, & DAC is equall the outward angle BCD (by the 32. of the 1.) Wherfore y^e angle BDA is equal vnto y^e angle BCD. But y^e angle BDA is (by y^e 5. of the first) equall vnto the angle CBD, for (by the 15. definition of y^• first) the side AD is equall vnto the side AB: wherfore (by the 1. common sentence) the angle DBA is equall vnto the angle BCD. Wherefore these three angles BDA, DBA, and BCD are equall the one to the other. And forasmuch as the angle DBC is equall vnto the angle BCD, the side therfore BD is equall vnto the side DC. But the line BD is by supposition equall vnto the lyne CA.

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two right angles. And either of the Angles at the base•• is two ••ift partes of two right angles, or foure fift partes of one right angle. Which shall manifestly appeare, if we deuide two right angles into fiue partes. For then in thys kinde of triangle, the angle at the toppe shall be one fift part, and eyther of the two angles at the base shall be two fift partes.

Thys also is to be noted, that the line AC is the side of an equilater Pentagon to be inscribed in the circle ACD. For by the latter construction it is manifest, that the three arkes AC, CD, and DE, of the lesse circle, are equall. And forasmuch as by the same it is manifest that the two lines AD and AE are equall, the arke also AE shall be equall to the arke AD (by the 20. of the third). Wherefore their halfes also are e∣quall. If therefore the arke AE be (by the 30. of the third) deuided into two equall partes, the whole circūference ACDEA shall be deuided into fiue equall arkes. And forasmuch as the lines subtending the sayd equall arkes are (by the 2••. of the same) e∣quall, therefore euery one of the sayd sides shall be the side of an equilater Pentagon•• which was required to be proued. And the same line AC shall be the side of an equila∣ter ten angled figure to be inscribed in the circle BDE: the demonstration wherof I omitte, for that it is demonstrated by Propositions following.

* 1.3¶ A Proposition added by Pelitarius.

Vpon a right line geuen being finite, to describe an equilater and equiangle Pentagon figure.

Suppose that the right line geuen be AB, vpon which it is required to describe an equilater and equiangle Pentagon. Vpon the line AB describe (by the 23. and 32. of the first) an Isosceles triangle ABC equiangle to the Isosceles triangle described by the former Proposition: namely, let the angles CAB and CBA, at the base AB, be equall to the two angles ABD and ADB in the former construction: so that eyther of them shall be two fift partes of two right angles, and the angle at the toppe, namely, the angle C, shall be one fift part. Then deuide the angle C into two equall parte•• by drawing the right line CD. And vpon the line AC, and vnto the poynt A, describe the angle CAD equall to the angle ACD, by drawing the line AD, which line AD let concurre with the line CD, in the poynt D•• and that within the triangle ABC, for the line CD being produced, shall fall vpon the base AB,

[illustration]

and the line AD vpon the side BC. And draw a line from the poynt D to the poynt B. And for asmuch as in the triangle ACD the two angles A and C are equall, ther∣fore (by the 6. of the first) the two sides AD and CD are equall. Againe forasmuch as the two sides CB and CD of the triangle CBD, are equall to the two sides CA and CD of the triangle ACD, and the angle C of the one, is equall to the angle C of the other (by construction) therefore (by the 4. of the first) the base DB is equall to the base DA, and so is equall to the line DC. Wherefore (by the 9. of the third) the poynt D shall be the centre of the circle described about the triangle ABC. Describe the sayd circle and let it be ABECF. Now then the an∣gle ADB is double to the angle ACD (by the 20. of the third). Wherefore the angle ADB maketh two fift partes of two right angles, that is, one fift part of foure right angles. And for asmuch as the space about the centre D is equall to foure right angles, then if the sayd space be deuided into fiue angles equall to the angle ADB, namely, into fine fift partes, by drawing the right lines DE & DF, which with the lines DA, DB, and DC, will cause the sayd space to be deuided into fiue equall partes, and if also there be drawen these right lines AF, FC, CE, and EB•• there shall be described a rectiline Pentagon figure A••ECF, which shall be equila∣••er, by the rule of a circle and of a circumference, and helpe of the 4. Proposition of the

first•• and shall also be equiangle (by the 4. and 5. of the same). For the fiue angles A, B, E, C, F, are deuided ech into ten equall partes: which was required to be done.

If we consider well thys demonstration of Pelitarius, it will not be hard for vs, vpon a right line geuen to describe the rest of the figures whose inscriptions here∣after followe.* 1.4

Notes

• * 1.1

  Construction.
• * 1.2

  Demonstra∣tion.
• * 1.3

  A Proposition added by Pe∣tarilius.
• * 1.4

  Note.