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 7. Probleme. The 12. Proposition. Vnto a right line geuen being infinite, and from a point geuen not being in the same line, to draw a perpendicular line.

LEt the right line geuen be∣ing

[illustration]

infinite be AB, & let y^e point geuen not being in the said line AB, be C. It is re∣quired from the point geuē, namely, C, to draw vnto the right line geuen AB, a per∣pendiculer line.* 1.1 Take on the other syde of the line AB (namely, on that syde wherein is not the pointe C) a pointe at all aduen∣tures, and let the same be D. And making the centre C, and the space CD, de∣scribe (by the third peticion) a circle, and let the same be EFG, which let cutte the line AB in the pointes E and G. And (by the x. proposition) deuide the lyne EG into two equal partes in the point H. And (by the first peticion) draw these right lines, CG, CH, and CE. Then I say, that vnto the right line geuen AB, & from the point geuen not being in it, namely, C, is drawen a perpendiculer lyne CH.* 1.2 For forasmuch as GH is equall to HE, and HC is common to them both; therfore these two sydes GH and HC, are equall to these two sydes EH & HC, the one to the other•• and (by the 15 definitiō) the base CG is equal to the base CE: wherfore (by the 8. proposition) the angle CHG is equall to the angle CHE: and they are syde angles: but when a right line standing vpon a right line maketh the two syde angles equall the one to the other, either of those equall an∣gles is (by the 10. definition) a right angle, and the line standing vpon the sayde right line is called a perpendiculer line. VVherfore vnto the right line geuē AB, and from the point geuen C, which is not in the line AB, is drawn a perpendicu∣ler line CH: which was required to be done.

This Probleme did Oen••pides first finde out,* 1.3 considering the necessary vse ther∣of to the study of Astronomy.

There ar•• two kindes of perpendiculer lines:* 1.4 wherof one is a plaine perpen∣diculer lyne, the other is a solide. A plaine perpendiculer line is, when the point from whence the perpendi••uler line i•• dra••en, is in the same plaine superficies with the line wherunto it is a perpendicular. A solide perpendiculer line is, whē the point, from whence the perpendiculer is drawne, is on high, and wi••hout the plaine superficies. So that a plaine perpendiculer line is drawen to a right line: & a solide perpendiculer line is drawn to a superficies. A plain•• perpendiculer line causeth right angles with one onely line, namely, with that vpon whome it fal∣leth. But a solide perpendiculer line causeth right an••le••, not only with one line, but with as many lynes as may be drawn in that superficies, by the touch therof.* 1.5 This proposition teacheth to draw a plaine perpendiculer line, for it is drawn to one line, and supposed to be in the selfe same plaine superficies.

There may be in this proposition an

[illustration]

other case.* 1.6 For if it be so, that on the o∣ther side of the line AB, there be no space to take a pointe in but onely on that side wherein is the point C.* 1.7 Then take some certaine point in the line AB, which let be D. And making the cen∣tre the point C, and the space CD, de∣scribe a part of the circumference of a circle, which let be DEF: which let cut the line AB in the two pointes D and F. And deuide the line DF into two e∣quall partes in the poynt H. And draw these lines CD, CH and CF. And for∣asmuch as DH is equal to HF,* 1.8 and CH is common to them both, and CD is equall to CF (by the 15. de∣finition:) therfore the angles at the point H are equal the one to the o∣ther (by the 8. proposition:) & they are side angles, wherefore they are right angles. Wherfore the line CH is a perpendiculer to the line DF. But if it happen so that the circle which is described do not cutte the lyne, but touche it, then takyng a point without the point E, name∣ly, the point G, and making the centre the point C, and the space CG, describe a part of the circumference of a circle: which shall of necessitie cut the line AB: and so may you proceede as you did before. As you see in the second figure.

Notes

• * 1.1

  Construction.
• * 1.2

  Demonstration.
• * 1.3

  Oenopides the first inuenter of this probleme.
• * 1.4

  Two kindes of perpendiculer lines, namely, a plaine perpendi∣culer line and a solide.
• * 1.5

  This proposition teacheth to draw a playne perpendiculer line.
• * 1.6

  An other case in this proposition.
• * 1.7

  Construction••
• * 1.8

  Demonstration.