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.
Page: 105 ([28])
[82.] LEMMA V.
Page: 106 ([29])
[83.] PROBLEM VII. (Fig. 32, 33, 34, &c.)
Page: 108 ([31])
[84.] PROBLEM I. (Fig. 32 to 45.)
Page: 109 ([32])
[85.] PROBLEM II. (Fig. 46 to 57.)
Page: 113 ([36])
[86.] PROBLEM III.
Page: 116 ([39])
[87.] THE END.
Page: 117 ([40])
[88.] A SYNOPSIS OF ALL THE DATA FOR THE Conſtruction of Triangles, FROM WHICH GEOMETRICAL SOLUTIONS Have hitherto been in Print.
Page: 133
[89.] By JOHN LAWSON, B. D. Rector of Swanscombe, in KENT. ROCHESTER:
Page: 133
[90.] MDCCLXXIII. [Price One Shilling.]
Page: 133
[91.] ADVERTISEMENT.
Page: 135
[92.] AN EXPLANATION OF THE SYMBOLS made uſe of in this SYNOPSIS.
Page: 137
[93.] INDEX OF THE Authors refered to in the SYNOPSIS.
Page: 139 ([i])
[94.] Lately was publiſhed by the ſame Author; [Price Six Shillings in Boards.]
Page: 140 ([ii])
[95.] SYNOPSIS.
Page: 141 ([1])
[96.] Continuation of the Synopsis, Containing ſuch Data as cannot readily be expreſſed by the Symbols before uſed without more words at length.
Page: 149 ([9])
[97.] SYNOPSIS
Page: 152 ([12])
[98.] FINIS.
Page: 156 ([16])
< >
page |< < ([10]) of 161 > >|
87[10]
In the indefinite line let the four points be A, E, I, U. It is then required
to be cut again in O ſo that OA x OU may be to OE x OI (be the Poſition
of the four given points what they may) in the ratio of AL to LE, let the
point L fall alſo as it may.
Construction. To the three points E, L, U, and the right line UI, let
be found by the IId Problem a fourth point O, ſo that UI x OE may be to
OU x OL as AE to AL. And let ſuch a Case be choſen of the IId Problem
that, according as UO is required greater or leſs than UI, or their ſum ſhall
conſtitute OI, ſo of the three rectangles deſcribed in the Vth Lemma made
by the four points E, O, A, L, that of OE x AL may accordingly be greater
or leſs than OL x AE, or their ſum conſtitute that of OA x EL.
N. B. U being now uſed to repreſent one of the given points, in all the
following Diagrams I have ſubſtituted V in the place where U was uſed
before.
Demonstration. On ſuppoſition therefore that ſuch a Case of the
IId Problem is made uſe of,
We have UI x OE: OU x OL: : AE: AL
And by Inverſion OU x OL: UI x OE: : AL: AE
And by Lemma IV. AL x OE: AE x OL: : OU: UI
Hence by Compoſition or Diviſion, & c. AL x OE: OA x LE: : OU
: OI as appears by Lemma V.
Then again by Lemma IV. OU x OA: OI x OE: : AL: LE
Q. E. D.
This Problem has three Epitagmas. The Iſt whereof is when of the
two aſſigned points A and U, the one of them is an extreme, and the other an
alternate mean; and this admits of three Cases. The IId is when A and U
are both of them extremes; and this has four Cases. The IIId is when of
A and U one of them is an extreme, and the other is the point next to it; and
this has three Caſes. And there remain three more Anomalous Caſes, wherein
the IId Problem is of no uſe, but which may be reduced to one, as ſhall be
ſhewn in it's proper place.
Epitagma I. Case I. Let A the firſt aſſigned point be an extreme, and
U the ſecond aſſigned point be an alternate mean; and let the point O be
ſought between the firſt aſſigned A and the next point to it E; or between
the ſecond aſſigned U and the laſt I. For the ſame Conſtruction ſerves
for both.

Text layer

  • Dictionary

Text normalization

  • Original

Search


  • Exact
  • All forms
  • Fulltext index
  • Morphological index
Impressum