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

An other demonstration after Campanus.

Suppose that there be a line AB, from whose ends A and B,

[illustration]

let there be drawen two lines AC and BC on one side, which let concur in the poynt C. Then I say that on the same side there can∣not be drawen two other lines, from the endes of the line AB, which shall concur at any other poynt, so that that which is drawē from the point A shall be equall to the line AC, and that which is drawen from the point B shalbe equall to the line BC. For if it be possible, let there be drawn two other lines on the selfe same side, which let concurre in the point D, and let the line AD be equall to the line AC, & the line BD equall to the line BC.* 1.1 Wherfore the poynt D shall fall either within the trian∣gle ABC, or without. For it cannot fall in one of the sides, for then a parte should be e∣quall to his whole. If therfore it fall without•• then either one of the lines AD and DB shall cut one of the lines AC and CB, or els neither shall cut neyther.* 1.2 Firste let one cut the other and draw a right line from C to D. Now forasmuch as in the triangle ACD, the two sides AC and AD are equall, therfore the angle ACD is equall to the angle ADC, by the fifth propositiō: likewise forasmuch as in the triāgle BCD, the two sides BC and BD are equall, therfore by the same, the angles BCD & BDC are also equall. And forasmuch as the angle BDC is greter thē the angle ADC,

[illustration]

it followeth that the angle BCD is greater then the angle ACD, namely, the part greater then the whole: which is impossible.

But if the point D fal without the triangle ABC,* 1.3 so that the lines cut not the one the other, draw a line from D to C. And produce the lines BD & BC beyond the base CD, vnto the points E & F. And forasmuch as the lines AC and AD are equall, the angles ACD and ADC shall also be equall, by the fifth proposition•• like∣wise for asmuch as the lines BC and BD are equal, the angles vn∣der the base, namely, the angles FDC and ECD are equall, by the seconde part of the same proposition. And for as much as the angle ECD is lesse then the angle ACD: It followeth that the angle FDC is lesse thē the angle ADC: which is impossible: for that the angle ADC is a part of the angle FDC. And the same inconuenience will follow if the poynt D fall within the triangle ABC,

Notes

• * 1.1

  Diuers cases in this demonstra∣tion.
• * 1.2

  First case.
• * 1.3

  Second case.