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 24. Theoreme. The 34 Proposition. In parallelogrammes, the sides and angles which are opposite the one to the other, are equall the one to the other, and their diameter deuideth them into two equall partes.

SVppose that ABCD be a parallelogramme and let the diameter ther•• of be BC. Then I say that the opposite sides and angles of the parallelo∣gramme ACDB are equall the one to the other, and y^t the diameter BC deuideth it into two equall partes.* 1.1 For foras∣much

[illustration]

as AB is a parallel line vnto CD, and v∣pon them falleth a right line BC: therfore (by the 29. proposition) the alternate angles ABC and BCD are equall the one to the other. A∣gayne forasmuch as AC is a parallel line to BD, and vppon them falleth the right lyne BC: therfore (by the same) the alternate angles, that is, the angles ACB and CBD are equall the one to the other. Now therfore there are two triangles ABC and BCD, hauing two an∣gles of the one, namely, the angles ABC and ACB equall to two angles of the other, that is, to the angles BCD and CBD, the one to the other, and one side of the one equal to one side of the other, namely, that syde that lieth betwene the equall angles, which syde is common to them both, namely, the side BC. VVher∣fore (by the 26. proposition) the other sides remaining are equall to the other sides remaining, the one to the other, and the angle remaining is equal to the an∣gle remayning. VVherfore the side AB is equall to the side CD, and the side AC

to the side BD, & the angle BAC is equal to the angle BDC. And for asmuch as the angle ABC is equal to the angle BCD, and the angle CBD to the an∣gle ACB: therfore (by the second common sentence) the whole angle ABD is equall to the whole angle ACD. And it is proued that the angle BAC is e∣quall to the angle CDB. VVherfore in parallelogrammes, the sides and angles which are opposite are equall the one to the other. I say also that the diameter therof deuideth it into two equall partes. For forasmuch as AB is equall to CD, and BC is common to them both, therfore these two AB and BC are equall to these two BC and CD, and the angle ABC is equal to the angle BCD. VVher¦fore (by the 4. proposition) the base AC is equall to the base BD, and the tri∣angle ABC is equall to the triangle BCD. VVherfore the diameter BC de∣uideth the parallelogramme ABCD into two equall partes: which is all that was required to be proued.

In this Theoreme•• are demonstrated three passions or properties of paralle∣logrammes.* 1.2 Namely, that thei•• opposite sides are equall: that their opposite an∣gles are equall: and that the diameter deuideth the parallelogramme into two e∣quall partes. VVhich is true in all kindes of parallelogrammes. There are fower kindes of parallelogrammes,* 1.3 a square, a ••igure of one side longer then the other, a Rhombus, or diamond figure, and a Rhom••oides or diamondlike figure. And here is to be noted, that in those parallelogrammes, all whose angles ar right an∣gles (as is a square, and a figure on the one side longer) the diameters do not only deuide the figure into two equall partes, but also they are equal the one to the o∣ther. As for example.

Suppose that ABCD be a square,

[illustration]

or a figure on the one side longer, and draw in it these diametres AD and BC. And forasmuch as the line AB is e∣quall to the line CD (by the definitiō of a square, and of a figure on the also one side lōger) & the line AC is com∣mon to the both: therfore two sides of the triangle ABC are equal to two sides of the triangle ACD, the one to the other, and the angles which they contayne are equall, namely, the an∣gles BAC & ACD, for they are rightangles. Wherefore the bases namely, the diame∣ters AD and BC, are (by the 4. proposition) equal.

But in those parallelogrames whose angles are not right angles, as is a Rhom∣bus, and a ••homboides, the diameters be euer vnequall. As for example.

Suppose that ABCD he

[illustration]

a Rhombus, or a Rhombaides and drawe in it these diame∣ters AC and BD. And foras∣much as AB is equall to CD and BC is common to them both, & the angle ABC is not equall to the angle BCD (by the definition of a Rhombus and also of a Rhombaides)

therefore (by the 24. proposition) the bases also are vnequall, namely, the diameters AC and BD.

Agayne. In parallelogrammes of equall sides, as are a square, and a Rhom∣bus, the diameters do not onely deuide the figures into two equall partes, but also they deuide the angles into two equall partes.

For suppose that there be a square or Rhombus ABCD, and draw the diameter AD. And forasmuch as the sides AB and BD are e∣quall

[illustration]

to the sides AC and CD (for the figures are equilater) and the angles ABD and ACD are equall (for they are opposite angles) and the base AD is common to both triangles. Therefore (by the fourth proposition) the angles BAD & CAD are equall, and so also are the angles BDA and CDA equall. Wh••••fore the angles BAC and CDB are deuided into two equall partes.

But in parallogrammes whose sides are not equall, such as are a figure on the one side longer, and a Rhomboides it is not so.

For suppose ABCD to be a figure on the one

[illustration]

side longer or a Romboides. And draw the dia∣meter AD. And now if the angles BAC and CDB, be deuided into two equall partes by the dia∣meter AD, then forasmuch as the angle. CAD is (by the 29. proposition) equall to the angle ADB, the angle also BAD shal be equal to the an∣gle ADB (by the first common sentence). Wher∣fore also the side AB is equall to the side BD (by the 6. propositiō). But the sayd sides are vnequal: which is impo••••ible. Wherefore the angles BAC and CDB are not deuided into two equall partes.

* 1.4The conuerse of the first and second part of this proposition after Proclus. Is 〈◊〉〈◊〉 figure whatsoeuer haue his opposite sides and angles equall: then is a parallelograms.

For suppose that ABCD be such a figure, namely, which hath his opposite sides and angles equall. And let the diameter thereof be AD.

[illustration]

Now forasmuch as the sides AB and BD are equall to to the sides DC and AC, and the angles which they cō∣tayne are equall, and the base AD is common to ech tri∣angle, therefore (by the 4. proposition) the angles remay∣ning are equall to the angles remayning, vnder which are subtended equal sides. Wherfore the angle BAD is equal to the angle ADC, and the angle ADB to the angle CAD. Wherefore (by the 27. proposition) the line AB is a parallel to the line CD, and the line AC to the line BD. Wherefore the figure ABCD is a parallogramme: which was required to be proued.

* 1.5A Corrollary taken out of Flussates.

A right line cutting a parallelogramme which way soeuer into two equall partes, shall also de∣uide the diameter thereof into two equall partes.

For if it be possible let the right line GC deuide the parallelogramme AEBD into two equal partes, but let it deuide the diameter DE into two vnequall partes in the point I. And l••t the part IE be greater then the part ID. And vnto the line ID put the line IO equall (by th3. propositiō).* 1.6 And by the

[illustration]

point O, draw vnto the lines AD and BE a pa∣rallel line OF (by the 31. proposition.) Where∣fore in the triangles FOI and CDI, two angles of the one are equal to two angles of the other, namely, the angles IOF and IDC (by the 29. proposition), & the angles FIO & CID (by the 15. propositiō), & the side ID is equal to the side IO. Wherefore (by the 26. proposition) the tri∣angles are equall. Wherefore the whole triangle EIG is greater then the triangle DIC. And forasmuch as the trapesium GBDC is sup∣posed to be the halfe of the parrallelograme, and the halfe of the same parallelograme is the triangle EBD (by this proposition). From the trapesium GBDC and the trian∣gle EBD which are equall, take away the trapesium GBDI which is common to them both, and the residue namely, the triangle DIC shalbe equall to the residue, namely, to the triāgle EIG: but it is also lesse (as hath before ben proued): which is impossible. Wherefore a right line deuiding a parallelogramme into two equall partes, shall not deuide the diameter thereof vnequally. Wherefore it shall deuide it equally: which was required to be proued.

An addition of Pelitarius.

Betwene two right lines being infinite and making an angle geuen:* 1.7 to place a line equall to a line geuen, in such sorte, that it shall make with one of those lines an angle equall to an other angle ge∣uen. Now it behoueth that the two angles geuen be lesse then two right angles.

Suppose that there be two lines AB and AC, making an angle geuen BAC: and let them be infinite on that side where they open one from the other. And let the line geuen be D, and let the other angle geuen be E.

[illustration]

And let the two angles A and E be lesse then two right angles (otherwise there coulde not be made a triangle, as it is manifest by the 17 proposition). It is required betwene the lines AB and AC to place a line equall to the line geuen D, which with one of them as for exāple with the line AC,* 1.8 may make an angle equal to the angle ge¦uen E. Now then vpon the line AC and to the point in it A, make an angle equall to the angle geuē E (by the 23. proposition), which let be CAF. And produce the line FA on the other side of the point A to the point G: and let AG be equall to the line geuen D (by the 3. proposition). And by the point G, draw (by the 31. proposition) a parallel line to the line AC, which let be GH, and produce it vntil it concurre with the line AB: which concurse let be in the point H. And agayne by the point H draw the line HK parallel vnto the line GF: which let cut the line AC in the point K. Then I say that the line HK is placed betwene the lines AB & AC & is equall to the line D. And that the angle at the point K is equall to the angle geuen E.* 1.9 For forasmuch as (by construction) AGHK is a parallelogramme the line KH is equall to the line AG (by this proposition). Wherefore also it is equall to the line D. And forasmuch as the line AK fall••th vppon the two parallel lines, FG and KH, therfore the angle AKH is equal to the angle FAK (by the 29. propositiō) for that they are alternate angles. Wherfore also the same angle at the point K is equal to the angle geuen E. Wherefore the line HK being placed betwene the two lines AB and AC and being equall to the line D•• maketh the angle at the point K equall to the angle geuen E: which was required to be done.

Though this addition of Pelitarius be not so muche pertayning to the

proposition: yet because it is witty and semeth somewhat difficult, I thought it good here to anexe it.

Notes

• * 1.1

  Demonstration
• * 1.2

  Thre passions of parallolegrames demōstrated in this Thoreme.
• * 1.3

  Fower kindes of parallelo∣grammes.
• * 1.4

  The conuerse of th•••• proposition after Pr••••lus.
• * 1.5

  A Corollary ta∣ken out of Flus∣sates.
• * 1.6

  Demonstration leading to 〈◊〉〈◊〉 absurdi••••e.
• * 1.7

  An addition o•• Pelitarius.
• * 1.8

  Construction.
• * 1.9

  Demonstration