Elements of geometry / from the Latin translation of Commandine. To which is added, a treatise of the nature of arithmetic of logarithms, likewise another of the elements of plain and spherical trigonometry; with a preface ... By Doctor John Keil ... The whole revised ... [with a] preface, by Samuel Cunn.

  • Euclid
Date:
1741
Catalogue details

Licence: Public Domain Mark

Credit: Elements of geometry / from the Latin translation of Commandine. To which is added, a treatise of the nature of arithmetic of logarithms, likewise another of the elements of plain and spherical trigonometry; with a preface ... By Doctor John Keil ... The whole revised ... [with a] preface, by Samuel Cunn. Source: Wellcome Collection.

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    t * Paft. I. ■j* I of this. t V°fi- * Pojl. 3. t Def 15. j Axiom 3. ^ a of this. t Poft‘ 3* J Axiom i r . J > Euclid's Elements. BookE Draw the Right Line AC from the Point A to C*, upon it defcribe the Equilateral Triangle DACf, produce D A and D C direftly forwards to E and G$; about the Centre C, with the Diftance B C, defcribe the Circle B G H *; and about the Centre Da With the Diftance D G, defcribe the Circle GKL. Mow becaufe the Point C is the Centre of the Circle BGH, BC will be equal to CGf; and becaufe D is the Centre of the Circle G K L, the Whole D L will be equal to the Whole DG, the Parts whereof DA and DC are equal; therefore the Remainders AL, GCj are alfo equal:}:. But it has been demon- ftrated, that B C is equal to CG; wherefore both AL and BC are each of them equal to CG. But Things that are equal to one and the fame Thing, are equal to one another; and therefore likewife A L is equal to BC. , . Whence the Right Line A L is put at the given Point A, equal to the given Right Line BC, which was to be done. PROPOSITION III. Problem. Two unequal Right Lines being given, to cut off d Part from the greater Equal to the leffer. * I ET AB and C be the two unequal Right Lines Tj given, the greater whereof is AB; it is required to cut off a Line from the greater A B equal to the leffer C. . . Put * a Right Line A D at the Point A, equal to the Line C; aqd about the Centre A, with the Diftance A D; defcribe a Circle D E Ff. Then, becaufe A is the Centre of the Circle DE F, AE is equal to AD; and fo both AE and C are each equal to AB; wherefore AE is likewife equal . to CR .. And fo there is cut off from AB,the greater of two given Right Lines AB and C, a Line AE equal to the leffer Line C; which was to be done.
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