Sinossi
This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1799 Excerpt: ...' KD DH will be greater than GD DH, and therefore KD is greater than GD, therefore KD2 is greater than GD2; but DG2=ADDB; wherefoie DK2 is greater than AD DB; therefore the ratio of AD CB: DK2 is less than the ratio of AD. CB: AD DB, that is, less than the ratio of CB: BD, but AD3:AC3::AD-CB:DK2; therefore the ratio of AD3: AC3 is less than the ratio of CB: BD. Q. E. D. Thesame otherwise by Mr. Lowry. On AB (fig. 131, pi. 9.) as a diameter describe the circle AFBP, and draw CFP perpendicular to AB: join AF, AP. By hypothesis AD3-4-AC3 is less than CB--BD; therefore AC3 CB is greaterthan AD3 DB; that is, AC3-CB must be a maximum. Now by the circle AC: CF:: CF: CB; therefore AC3: AC 2 CF: CF: CB; therefore AC 3 CB=AC2 CF2; wherefore AC2-CF2, or AC'CF, or the triangle APF must be a maximum. But the greatest triangle that can be inscribed in a circle, is well known to be the equilateral one; therefore AC=5 AB=3BC; consequently the truth of the proposition is manifest. XVIII. v (x--j)--x--yz will be the fluxion required. This question was also answered by Messrs. Lowry and Simpson. XX. QUESTION 28, will be answered in No. IV. ARTICLE XXXI. Four Propositions from Lawson. (To be answered in Number V.) PROP. XVIII. Let ABC be a triangle inscribed in a circle, whose sides AB and AC are equal, and from A any line be drawn meeting the circle again in D and BC in E; I fay that the rectangle DAE is equal to the square of AB. PROP. XIX. Things remaining as in the last proposition, of lines touching the circle in A and C be drawn to meet in F and FD be drawn cutting BC in G; I fay that the rectangle BCG is equal to the square ofCE. PROP. XX. Let ABC be a triangle inscribed in a circle, whose sides AB and AC are equal, and let AD be parallel to BC, and taking any point D t...
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