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 18. Theoreme. The 24. Proposition. In euery parallelogramme, the parallelogrammes about the dimecient are lyke vnto the whole, and also lyke the one to the other.

* 1.1SVppose y^t there be a parallelogramme ABCD, and let the dimecient therof be AC: and let the parallelogrammes about the dimecient AC, be EG and HK. Then I say that either of these parallelogrames EG and HK is like vnto the whole parallelogramme ABCD, and also are lyke the one to the other. For forasmuch as to one of the sides of the trian∣gle ABC, namely, to BC is drawen a parallel lyne EF, therfore as BE is to EA, so (by the 2. of the sixt) is CF to FA. Agayne forasmuch as to one of y^e sides

of the triangle ADC, namely, to CD is drawen a parallel lyne F••, therefore (by the same) as CF is to FA, so is DG to GA. But as CF is to FA, so is it pro••ued that BE is to EA. Whe••fore as BE is to EA, so (by the 11. of the fifth)•• is DG to GA. Wherfore by composition (by the 18. of the fifth) as BA is to AE•• so is DA to AG. And alternately (by the 16. of the fifth) as BA is to AD, so is EA to AG. Wherfore in the parallelogrammes•• ABCD and EG y^e sides which are about the common angle BAD are proportionall. And because y^e line GF is a parallel vnto the lyne DC•• therfore the angle AGF (by the 29•• of the•• first) is equall vnto y^e angle ADC, •• y^e

[illustration]

angle GFA equall vnto y^e angle DCA and the angle DAC is common to the two triangles ADC and AFG: Wher∣fore the triangle DAC is equiangle vnto the triangle AGF. And by the same reason the triangle ABC is equi∣angle vnto the triangle AEF. Wher∣fore the whole parallelogramme ABCD is equiangle vnto the parallelogrāme EG. Wherfore as AD is in proportion to DC, so (by the 4. of the sixth) is AG to GF, and as DC is to CA, so is GF to FA. And as AC is to CB, so is AF to FE. And moreouer as CB is to BA, so is FE to EA. And forasmuch as it is proued that as D•• is to CA, so is GF to FA: but as AC is to C••, so is AF to FE. Wherfore of equalitie (by the 22. of the fifth) as DC is to CB, so is GF to FE. Wherefore in the parallelogrammes ABCD and EG, the sides which include the equall angles are proportionall. Wherefore the parallelogramme ABCD is (by the first definition of the sixth) like vnto the parallelogramme EG.

And by the same reason also the parallelogramme ABCD is like to the pa∣rallelogramme KH:* 1.2 wherefore either of these parallelogrammes EG and KH is like vnto the parallelogramme ABCD. But rectiline figures which are like to one and the same rectiline figure are also (by the 21. of the sixth) like the one to the other. Wherefore the parallelogramme EG is like to the parallelo∣gramme HK.* 1.3 Wherfore in euery parallelogramme, the parallelogrammes a∣bout the dimecient are like vnto the whole, and also like the one to the other. Which was required to be proued.

¶ An other more briefe demonstration after Flussates.

Suppose that there be a parallelogrāme ABCD, whose dime••ient let b•• A••,* 1.4 about which let consist these parallelogrammes EK and TI, hauing the angles at the pointes •• and 〈…〉〈…〉 with the whole parallelogramme ABCD. Then I say, that those pa∣rallelogrammes EK and TI are like to the whole parallelogramme DB and also al••

like the one to the other. For forasmuch as BD, EK,

[illustration]

and TI are parallelogrammes, therefore the right line AZG falling vpon these parallell lines AEB, KZT, and DI G, or vpon these parallell lines AKD, EZI, and BTG, maketh these angles equall the one to the other, namely, the angle EAZ to the angle KZA, & the an∣gle EZA to the angle KAZ, and the angle TZG to the angle ZGI, and the angle TGZ to the angle IZG, and the angle BAG to the angle AGD: and finally, the angle BGA to the angle DAG. Wherefore (by the first Corollary of the 32. of the first, and by the 34. of the first) the angles remayning are equall the one to the other, namely, the angle B to the angle D, and the angle E to the angle K, and the angle T to the angle I. Wherefore these triangles are equiangle and therefore like the one to the other, namely, the triangle ABG to the triangle GDA, and the triangle AEZ to the triangle ZKA, & the triangle ZTG to the triangle GIZ. Wherefore as the side AB is to the side BG, so is the side AE to the side EZ, and the side ZT to the side TG. Wherefore the parallelogrammes contayned vnder those right lines, namely, the parallelogrammes ABGD, EK, & TI, are like the one to the other (by the first definition of this booke). Wherefore in euery parallelogramme the parallelogrammes. &c. as before: which was required to be de∣monstrated.

¶ A Probleme added by Pelitarius.

Two equiangle Parallelogrammes being geuen, so that they be not like, to cut of from one of them a parallelogramme like vnto the other.

* 1.5Suppose that the two equiangle parallelogrammes be ABCD and CEFG, which let not be like the one to the other. It is required from the Parallelogramme ABCD, to cut of a parallelogramme like vnto the parallelogramme CEFG. Let the angle C of the one be equall to the angle C of the other. And let the two parallelogrammes be so 〈◊〉〈◊〉, that the lines BC & CG may make both

[illustration]

one right line, namely, BG. Wherefore also the right lines DC and CE shall both make one right line, namely, DE. And drawe a line from the poynt F to the poynt C, and produce the line FC till it cōcurre with the line AD in the poynt H. And draw the line HK parallell to the line CD (by the 31. of the first). Then I say, that from the paralle∣logramme AC is cut of the parallelogrāme CDHK, like vnto the parallelogrāme EG. Which thing is manifest by thys 24. Propo∣sition. For that both the sayd parallelo∣grammes are described about one & the selfe same dimetient. And to the end it might the more plainly be seene, I haue made complete the Parallelogramme ABGL.

¶ An other Probleme added by Pelitarius.

Betwene two rectiline Superficieces, to finde out a meane superficies proportionall.

* 1.6Suppose that the two superficieces be A and B, betwene which it is required to place a meane superficies proportionall. Reduce the sayd two rectiline figures A and B

vnto two like parallelogrāmes (by the 18. of this booke) or if you thinke good reduce eyther of them to a square, (by the last of the second). And let the said two parallelo∣grammes like the one to the other and equall to the superficieces A and B, be CDEF and FGHK. And let the angles F in either of them be equall, which two angles let be placed in such sort, that the two parallelogrammes ED and HG may be about one and the selfe same dimetient CK (which is done by putting the right lines EF and FG in such sort that they both make one right line, namely,

[illustration]

EG). And make cōplete the parallelogrāme CLK M. Then I say, that either of the supplements FL & FM is a meane proportionall betwene the superficieces CF & FK, that is, betwene the superficieces A and B: name∣ly, as the superficies HG is to the superficies FL, so is the same superficies FL to the superficies ED. For by this 24. Proposition the line HF is to the line FD, as the line GF is to the line FE. But (by the first of this booke) as the line HF is to the line FD, so is the su∣perficies HG to the superficies FL: and as the line GF is to the line FE, so also (by the same) is the superficies FL to the superficies ED. Wherfore (by the 11. of the fift) as the superficies HG is to the superficies FL, so is the same superficies FL to the superficies ED: which was required to be done.

Notes

• * 1.1

  Demonstration of this propositiō wherein is first proued that the parallegramme EG is like to the whole parallelo∣grāme ABCD.
• * 1.2

  That the paral∣lelogrāme KH is like to the whole parallelo∣gramme AB∣CD.
• * 1.3

  That the paral∣lelogrammes EG and KH are like the one to the other.
• * 1.4

  An other Demonstra∣tion after Flussates.
• * 1.5

  An addition of Pelitarius.
• * 1.6

  Another ad∣dition of Pelitarius.