Apollonius <Pergaeus>; Lawson, John, The two books of Apollonius Pergaeus, concerning tangencies, as they have been restored by Franciscus Vieta and Marinus Ghetaldus : with a supplement to which is now added, a second supplement, being Mons. Fermat's Treatise on spherical tangencies

Table of contents

< >
[81.] LEMMA IV.
[82.] LEMMA V.
[83.] PROBLEM VII. (Fig. 32, 33, 34, &c.)
[84.] PROBLEM I. (Fig. 32 to 45.)
[85.] PROBLEM II. (Fig. 46 to 57.)
[86.] PROBLEM III.
[87.] THE END.
[88.] A SYNOPSIS OF ALL THE DATA FOR THE Conſtruction of Triangles, FROM WHICH GEOMETRICAL SOLUTIONS Have hitherto been in Print.
[89.] By JOHN LAWSON, B. D. Rector of Swanscombe, in KENT. ROCHESTER:
[90.] MDCCLXXIII. [Price One Shilling.]
[91.] ADVERTISEMENT.
[92.] AN EXPLANATION OF THE SYMBOLS made uſe of in this SYNOPSIS.
[93.] INDEX OF THE Authors refered to in the SYNOPSIS.
[94.] Lately was publiſhed by the ſame Author; [Price Six Shillings in Boards.]
[95.] SYNOPSIS.
[96.] Continuation of the Synopsis, Containing ſuch Data as cannot readily be expreſſed by the Symbols before uſed without more words at length.
[97.] SYNOPSIS
[98.] FINIS.
< >
page |< < ([6]) of 161 > >|
83[6]
Therefore AT x AO is greater than o AO
Or AT greater than Ao.
Corollary II. If the three given points be I, A, E; and O falls between
A and I, ſo as to make AO x PE:
IOE: : AL: LI, I ſay then O will fall
beyond L.
For let us ſuppoſe that O and L coincide; then by hypotbeſis AL: LI: :
AL x PE:
IL x LE
And by the next following Lemma IV. AL x IL: IL x PE: : AL: LE
i.
e. AL: PE: : AL: LE
Hence PE is equal to LE, a part to the whole, which is manifeſtly abſurd.
LEMMA IV.
If it be as a line to a line ſo a rectangle to a rectangle; then I ſay it will be
as the flrſt line into the breadth of the ſecond rectangle to the ſecond line into
the breadth of the firſt rectangle, ſo the length of the firſt rectangle to the
length of the ſecond.
Suppoſition. AE: IO: : UYN: SRL.
Concluſion. AE x RL: IO x YN: : UY: SR.
Dem. AE: IO: : AE x YN: IO x YN: : UYN: SRL
And by Permutation AE x YN: UYN: : AE: UY: : IO x YN: SRL
But SR: AE: : SRL: : AE x RL
Therefore ex æquo perturbatè SR: UY: : IO x YN: AE x RL
Q. E. D.
LEMMA V.
If a right line be cut in two points, I fay the rectangle under the alternate
ſegments is equal to that under the whole and the middle ſegment, together
with the rectangle under the extremes.
Dem. AI x IE + IO x IE = AO x IE.
Hence {AI x IE + IO x IE + AE x IO \\ i. e. AI x IE + AI x IO \\ i. e. AI x EO} = AO x IE + AE x IO.
Q. E. D.
N. B. Theſe two Lemmas ſave much Circumlocution and Tautology in
the two following Propoſitions, and indeed are highly uſeful in all caſes where
compound ratios are concerned.

Text layer

  • Dictionary

Text normalization

  • Original
  • Regularized
  • Normalized

Search


  • Exact
  • All forms
  • Fulltext index
  • Morphological index