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 13. Probleme. The 45 Proposition. To describe a parallelograme equal to any rectiline figure ge∣uen, and contayning an angle equall to a rectiline angle geuē.

SVppose that the rectiline figure geuen be ABCD, and let the rectiline angle geuē be E. It is required to describe a parallelograme equall to the rectiline figure geuen ABCD, and contayning an angle equal to the re∣ctiline angle geuen E.* 1.1 Draw (by the first peticion) a right line frō the point D to the point B. And (by the 42. proposition) vnto the triangle ABD describe an e∣quall parallelograme FH, hauing his angle FKH equall to the angle E. And (by the 4••. of the first) vpō the right

[illustration]

line GH apply the parallelogramme GM equal to the triangle DBC, ha∣uing his angle GHM equall to the angle E.* 1.2 And forasmuch as eyther of those angles HKF and GHM is equall to the angle E: therefore the angle HKF is equall to the angle GHM: put the angle KHG com∣mon to them both, wherfore the an∣gles FKH and KHG are equall to the angles KHG and GHM. but the angles FKH and KHG are (by the 29. proposition) equall to two right angles. VVherfore the angles KHG and GHM are equall to two

right angles. Now then vnto a right line GH, and to a point in the same H, are drawen two right lines KH and HM not both on one and the same side, ma∣king the side angles equall to two right angles. VVherfore (by the 14. propositi∣on) the lines KH and HM make directly one right line. And forasmuch as v∣pon the parallel lines. KM and FG falleth the right line HG, therefore the alternate aagles MHG and HGF are (by the 29 proposition) equall the one to the other: put the angle HGL common to them both. VVherfore the angles MHG and HGL are equall to the angles HGF and HGL. But the angles MHG & HGL are equall to two right angles (by y^e 29. proposition). VVher∣fore also the angles HGF and HGL are equall to two right angles. VVher∣fore (by the 14. proposition) the lines FG and GL make directly one right line. And forasmuch as the line KF is (by the 34. proposition) equal to the lyne HG, and it is also parallel vnto it: and the line HG is (by the same) equall to the line ML, therfore (by the first common sentence) the line FK is equall to the lyne ML, and also a parallel vnto it (by the 30. proposition). But the right lynes KM and FL ioyne them together. VVherfore (by the 33. proposition) the lines KM and FL are equall the on to the other and parallel lines. VVherfore KFLM is a parallelograme. And forasmuch as the triangle ABD is eqnal to the paral∣lelograme FH, and the triangle DBC to the parallelogramme GM•• therfore the whole rectiline figure ABCD is equall to the whole parallelograme KFLM. VVherfore to the rectiline figure geuen ABCD is made an equall paralle∣grame KFLM, whose angle FKM is equal to the angle geuen, namely, to E: which was required to be done.

The rectiline figure geuē 〈◊〉〈◊〉 in the example of Euclide is a parallelograme. But if the rectiline figure be of many sides, as of 5.6. or mo, thē must you resolue the figure into his triangles, as hath bene before taught in the 32. proposition. And thē apply a parallelograme equal to euery triangle vpon a line geuē, as before in the example of the author. And the same kind of reasoning wil serue that was be¦fore, only by reasō of the multitude of triangles, you shall haue neede of oftener repeticiō of the 29. and 14. propositiōs to proue that the bases of al the parallelo∣grames made equall to all the triangles make one right line, and so also of the toppes of the said parallelogrames. Pelitarius addeth vnto this proposition this Probleme following.

Two vnequall rectiline superfici••c••s beyng geuen, to find out the exc••sse of the greater aboue the lesse.* 1.3

Suppose that

[illustration]

there be two vne∣quall rectiline su∣perficieces A & B of which let A be the greater. It is required to finde out the excesse of the superficies A aboue the super∣ficieces B. De∣scribe (by the 44.

proposition) the parallelograme CDEF equall to the rectiline figure A, contayning a right angle. And produce the line CD beyond the point D to the point G: & put the line DG equall to the line CD. And againe (by the 44. proposition) vpon the line DG describe the parallelograme DGHK equall to the rectiline figure B, and hauyng the angle DGK a right angle. And produce the line KH beyond the point H, vntill it cutte the line CE in the point L. Then I say that HLEF, is the excesse of the rectiline figure A aboue the rectiline figure B. For first that CGKL is a parallelogramme it is mani∣fest, neither nedeth it to be demonstrated. And forasmuch as the lines CD and DG are by supposition equal and either of them is a parallel to KL, therfore (by the 36, propo∣sition) the two parallelogrames CH and DK are equall. And forasmuch as DK is sup∣posed to be equall to the rectiline figure B, CH also shall be equall to the same rectiline figure B. Wherfore forasmuch as the whole parallelograme CF is equall to the rectiline figure A, and LF is the excesse of CF aboue DL or DK, it followeth that LF is the ex∣cesse of the rectiline figure A aboue the rectiline figure B: whiche was required to be done.

An other more redy way.

Let the parallelograme CDEF remayne equall to the rectiline figure A, & produce the line CD beyond the point D to the pointe G. And vpon the line DG describe the parallelograme DGHK equall to the rectiline figure B. And produce the lines EC & HK beyond the points C and K till they concurre in the point L. And by the pointe D draw the diame∣ter

[illustration]

LDM, which let cutte the line HG beyng pro∣duced beyonde the pointe G in the point M, & by the pointe M drawe vnto the line HL a paral∣lel MN cuttyng the line EL in the pointe N: and by that meanes is HLMN a paralle∣lograme. Then I say that NF is the evcesse of the rectiline figure A aboue the rectiline figure B. For forasmuch as the parallelograme HD is equall to the rectiline figure B, & the supplementes HD and DN ar (by the 43. proposition) equall: therfore DN also is equall to the rectiline figure B, which rectiline figure DN being taken away frō the parallelograme CF (which is supposed to be equall to the rectiline figure A) the resi∣due NF shall be the ezcesse of the rectiline figure A aboue the rectiline figure B: which was required to be done.

Notes

• * 1.1

  Construction.
• * 1.2

  Demonstration.
• * 1.3

  An addition of Pelitarius.