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

[51.] PROBLEM X.

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[52.] PROBLEM XI.

[53.] PROBLEM XII.

[54.] PROBLEM XIII.

[55.] PROBLEM XIV.

[56.] PROBLEM XV.

[57.] Synopſis of the PROBLEMS.

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[58.] THE TWO BOOKS OF APOLLONIUS PERGÆUS, CONCERNING DETERMINATE SECTION, As they have been Reſtored by WILLEBRORDUS SNELLIUS. By JOHN LAWSON, B. D. Rector of Swanſcombe, Kent. TO WHICH ARE ADDED, THE SAME TWO BOOKS, BY WILLIAM WALES, BEING AN ENTIRE NEW WORK. LONDON: Printed by G. BIGG, Succeſſor to D. LEACH. And ſold by B. White, in Fleet-Street; L. Davis, in Holborne; J. Nourse, in the Strand; and T. Payne, near the Mews-Gate. MDCC LXXII.

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[59.] ADVERTISEMENT.

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[60.] EXTRACT from PAPPUS's Preface to his Seventh Book in Dr. HALLEY's Tranſlation. DE SECTIONE DETERMINATA II.

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[61.] THE PREFACE.

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[62.] PROBLEMS CONCERNING DETERMINATE SECTION. PROBLEM I.

[63.] LEMMA I.

[64.] LEMMA II.

[65.] LEMMA III.

[66.] PROBLEM II.

[67.] LEMMA IV.

[68.] LEMMA V.

[69.] PROBLEM III.

[70.] PROBLEM IV.

[71.] DETERMINATE SECTION. BOOK I. PROBLEM I. (Fig. 1.)

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[72.] PROBLEM II. (Fig. 2 and 3.)

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[73.] PROBLEM III. (Fig. 4. and 5.)

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[74.] PROBLEM IV. (Fig. 6. 7. and 8.)

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[75.] PROBLEM V. (Fig. 9. 10. 11. 12. 13. 14. 15. 16.)

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[76.] PROBLEM VI. (Fig. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26.)

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[77.] THE END OF BOOK I.

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[78.] DETERMINATE SECTION. BOOK II. LEMMA I.

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[79.] LEMMA II.

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[80.] LEMMA III.

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80[3]

LEMMA II.

If of four proportionals the ſum of two, being either extremes or means, be
greater than the ſum of the other two; then I ſay theſe will be greateſt and
leaſt of all.

This is the converſe of Euc. V. 25. and may be thus demonſtrated. Draw
a circle whoſe diameter may be equal to the greater ſum; and in it inſcribe the
leſſer ſum IO, which will therefore not paſs through the center, and let the
parts be IU and UO; then through U draw a diameter AUE, and the other
two terms will be AU and EU, of which AU is greateſt of all and EU leaſt of
all, and IU and UO of intermediate magnitude, by Euc. III. 7.

LEMMA III.

If of four proportionals the difference of two, being either extremes or
means, be greater than the difference of the other two, then I ſay theſe will be
the greateſt and leaſt of all.

This is demonſtrated in the ſame manner as the preceding by Euc. III. 8.

PROBLEM II.

To cut a given indefinite right line in one point, ſo that, of the three ſeg-
ments intercepted between the ſaid point and three points given in the ſame in-
definite right line, the rectangle under one of them and a given external right
line may be to the rectangle under the other two in a given ratio.

In the given indefinite line let the aſſigned points be A, E, I. It is then
required to cut it again in the point O, ſo that AO into a given external line
R may be to EO x IO as R to S. If A be an extreme point and E the
middle one, then ſet off IU = AE the contrary way from A; but if A be the
middle point, then ſet it off towards A. Then from U ſet off UN = S the
conſequent of the given ratio, either towards A, or the contrary way; for as
the Caſes vary, it’s poſition will vary. From A and N erect perpendiculars
AY and NM to the given indefinite right line equal to AE and AI re-
ſpectively, and theſe bomotactical if A be an extreme point, but antitactical if
A be the middle point of the three given ones. Join the extremes of theſe
perpendiculars Y and M, and upon YM as a Diameter deſcribe a circle. I ſay
that the interſection of this circle with the given indefinite line ſolves the
Problem. If it interſects the line in two points, then the Problem admits

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