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 5. Theoreme. The 5. Proposition. If two triangles haue their sides proportionall, the triang••••s are equiangle, and those angles in thē are equall, vnder which are subtended sides of like proportion.

SVppose that there be two triangles ABC, & DEF, hauing their sides proportionall, as AB is to BC, so let DE be to EF: & as BC is to AC, so let EF be to DF: and moreouer, as BA is to AC, so let ED be to DF. Then I say, that the triangle ABC is equiangle vnto the triangle DEF: and those angles in them are equall vnder which are subtended sides of like pro∣portion, that is, the angle ABC is equall vnto the angle DEF: and the angle BCA vnto the angle EFD: and moreouer, the angle BAC to y^e angle EDF. Vpon the right line EF, and vnto the pointes in it E & F, describe (by the 23. of the first) angles equall vnto the angles ABC & ACB, which let be FEG and EFG, namely, let the angle FEG be equall vnto the angle ABC, and let the angle EFG be equall to the angle ACB. And forasmuch as the angles ABC and ACB are lesse then two right angles (by the 17. of the first): therefore also the angles FEG and EFG are lesse then two right angles. Wherefore (by the 5. petition of y^e first) y^e right lines EG & FG shall at y^e length concurre. Let thē concurre in the poynt G. Wherefore EFG is a triangle. Wherefore the angle remayning BAC is equall vnto the angle remay∣ning

[illustration]

EGF (by the first Corollary of the 32. of the first). Wherfore the triangle ABC is equiangle vn∣to the triangle GEF. Wherefore in the triangles ABC and EGF the sides, which include the equall angles (by the 4. of the sixt) are proportionall, and the sides which are subtended vnder the equall an∣gles are of like proportion. Wherefore as AB is to BC, so is GE to EF. But as AB is to BC, so by supposition is DE to EF. Wherefore as DE is to EF, so is GE to EF (by the 11. of the fift). Where∣fore either of these DE and EG haue to EF one and the same proportion. Wherefore (by the 9. of the fift) DE is equall vnto EG. And by the same rea∣son also DF is equall vnto FG. Now forasmuch as DE is equall to EG and EF is common vnto them both, therefore these two sides DE & EF, are equall vnto these two sides GE and EF, and y^e base DF is equall vnto the base FG. Wherefore the angle DEF (by the 8. of the first) is equall vnto the angle GEF: and the triangle DEF (by the 4. of the first) is equall vnto the triangle GEF: and the rest of the angles of the one triangle

are equall vnto the rest of the angles of the other triangle the one to y^e other, vn∣der which are subtended equall sides. Wherefore the angle DFE is equall vn∣to the angle GFE: and the angle EDF vnto the angle EGF. And because•• the angle FED is equall vnto the angle GEF: but the angle GEF is equall vnto the angle ABC: therefore the angle ABC is also equall vnto the angle FED. And by the same reason the angle ACB is equall vnto y^e angle DFE•• and moreouer, the angle BAC vnto the angle EDF. Wherefore the triangle ABC is equiangle vnto the triangle DEF. If two triangles therefore haue their sides proportionall, the triangles shall be equiangle, & those angles in them shall be equall, vnder which are subtended sides of like proportion: which was required to be demonstrated.

Notes

• This is the conuerse of the former proposition.
• Construction
• Demonstra∣tion.