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 17. Theoreme. The 26. Proposition. If two triangles haue two angles of the one equall to two an∣gles of the other, ech to his correspondent angle, and haue also one side of the one equall to one side of the other, either that side which lieth betwene the equall angles, or that which is subtended vnder one of the equall angles: the other sides also of the one, shalbe equall to the other sides of the other, eche to his correspondent side, and the other angle of the one shalbe equall to the other angle of the other.

SVppose that there be two triangles AB

[illustration]

C, and DEF, hauing two angles of the one, that is, the angles ABC, and BCA, equall to two angles of the other, that is, to the angles DEF, and EFD, ech to his correspō¦dent angle, that is, the angle ABC, to the angle DEF, and the angle BCA to the angle EFD, and one side of the one equall to one side of y^e other, first that side which lieth betwene the equall angles, that is, the side BC, to the EF. Thē I say that the other sides also of the one shalbe equall to the other sides of the other, ech to his corre∣spondent side, that is, that side AB, to the side DE, and the side AC, to the side DF, and the other angle of the one, to the other angle of the other, that is, the angle BAC to the angle EDF. For if the side AB be not equall to the side DE, the one of them is greater. Let the syde AB be greater: and (by the 3. propo¦sition) vnto the line DE, put an equall line GB, and draw a right line from the point G, to the point C. Now forasmuch as the line GB,* 1.1 is equall to the line DE, and the line BC to the line EF, therefore these two lines GB and BC, are equall to these two lines DE and EF, the one to the other, and the angle GBC is (by supposition) equall to the angle DEF. VVherefore (by the 4. proposy∣tion) the base GC is equall to the base DF, and the triangle GCB is equall to the triangle DEF, and the angles remay••ing are equall to the angles remay∣ning vnder which are subtended equall sydes. VVherefore the angle GCB is e∣quall to the angle DFE. But the angle DFE is supposed to be equall to the an¦gle BCA. VVherefore (by the first common sentence) the angle BCG is equal to the angle BCA, the lesse angle to the greater: which is impossible. VVhere∣fore the line AB is not vnequall to the line DE. VVherefore it is equall And the the line BC is equall to the line EF: now therefore there are two sydes AB and

BC equall to two sydes DE and EF, the one to the other, and the angle ABC, is equall to the angle DEF. VVherefore (by the 4. proposition) the base AC is equall to the base DF, and the angle remayning BAC is equall to the angle re¦mayning EDF.

Agayne suppose that the sydes subtending the equall angles be equall the one to the other, let the syde I say AB be equall to the syde DE. Then agayne I say, that the other sydes of the one are equall to the other sydes of the other, ech to his correspondent syde, that is the syde AC to the syde DF, and the syde BC to the syde EF: and moreouer the angle remayning, namely, BAC, is equall to the angle remayning, that is, to the angle EDF. For if the syde BC be not equall to the syde EF, the one of them is greater: let the syde BC, if it be possi∣ble, be greater. And (by the third proposytion) vnto the line EF, put an equall line BH, and drawe a right line from the point A to the point H. And forasmuch as the line BH is equall to the line EF, and the line AB to the line DE, therefore these two sydes AB and BH, are equall to these two sydes DE and EF, the one to the other, and they containe

[illustration]

equall angles. VVherefore (by the 4. proposition) the base AH is equall to the base DF, and the triangle ABH, is equall to the triangle DEF, and the an∣gles remayning are equall to the angle•• remayning, vnder which ar subtēded equal sydes. VVherfore the angle BHA is equall to the angle EFD. But the angle EFD is equall to the angle BCA. VVhere∣fore the angle BHA is equal to the angle BCA. VVherefore the outward angle of y^e triangle AHC, namely, the angle BHA, is equall to the inward and opposite angle, namely, to the angle HCA, which (by the 16 proposition) is impossible. VVherfore the syde EF is not vnequall to the syde BC, wherefore it is equall. And the syde AB is equall to y^e syde DE: wherefore these two sydes AB and BC, are equall to these two sydes DE and EF, the one to the other, and they contayne equall angles: VVherfore (by the 4. proposition) the base AC is equall to the base DF: and the triangle ABC, is equall to the triangle DEF, and the angle remayning, name∣ly, the angle BAC is equall to the angle remayning, that is, to the angle EDF. If therefore two triangles haue two angles of the one equall to two angles of the other, ech to his correspondent angle, and haue also one syde of the one equall to o••e syde of the other, either that syde which lieth betwene the equall angles, or that which is subtended vnder one of the equall angles: the other sydes also of the one shalbe equall to the other sydes of the other, eche to his correspondent side, and the other angle of the one shalbe equall to the other angle of the other: which was required to be proued.

VVhereas in this proposition it is sayde, that triangles are equall, which hauing two angles of the one equall to two angles of the other, the one to the o∣ther,

haue also one side of the one equall to one side of the other, either that side which lieth betwene the equall angles, or that side which subtendeth one of the equall angles: this is to be noted that without that caution touching the equall side, the proposition shall not alwayes be true. As for example.

Suppose that there be a rectangle triangle ABC, whose right angle let be at the point B, & let the side BC be greater thē the side BA: and produce the line AB, frō the point B to the point D. And vpō the right line

[illustration]

BC & to the point in it C, make vnto the angle BAC an equal angle (by the 23. proposition), which let be BCD, & let the lines BD & CD, be¦ing produced cōcurre in the point D. Now thē there are two triangles ABC, and BCD, which haue two angles of the one equall to two an∣gles of the other, the one to the other, namely, the angle ABC to the angle DBC (for they are both right angles), & the angle BAC, to the angle BCD (by construction) and haue al∣so one side of the one equall to one side of the other, namely, the side BC, which is cō∣mon to them both. And yet notwithstanding the triangles are not equall: for the tri∣angle BDC, is greater then the triangle ABC. For vpon the right line BC, and to the point in it C, describe an angle equall to the angle ACB: which let be FCB (by the 23. proposition). And forasmuch as the side BC was supposed to be greater then the side. AB, therefore (by the 18. proposition) the angle BAC is greater then the angle BCA, wherefore also the angle BCD is greater then the angle BCF. Wherefore the tri∣angle BCD is greater then the triangle BCF. Agayne forasmuch as there are two tri∣angles ABC and BCF; hauing two angles of the one equal to two angles of the other•• the one to the other, namely, the angle ABC to the angle FBC (for they are both right angles) and the angle ACB to the angle FCB (by construction), and one side of the one is equall to one side of the other, namely, that side which lieth betwene the equall an∣gles, that is, the side BC which is common to both triangles. Wherefore (by this pro∣position) the triangles ABC and FBC are equal. But the triangle DBC is greater thē the triangle FBC. Wherefore also the triangle DBC is greater then the triangle ABC. Wherefore the triangles ABC and DBC, are not equall: notwithstanding they haue two angles of the one equall to two angles of the other, the one to the other, and one side of the one equall to one side of the other.

The reason wherof is, for that the equal side in one triangle, subtēdeth one of the equall angles, and in the other lieth betwene the equal angles. So that you see that it is of necessitie that the equall side do in both triangles, either subtend one of the equall angles, or lie betwene the equall angles.

Of this proposition was Thales Milesius the inuentor,* 1.2 as witnesseth Eude∣mus in his booke of Geometricall enarrations.

Notes

• * 1.1

  Demonstration leading to an absurditis.
• * 1.2

  Thales Milesius the inuent••r of this proposition.