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.

    126/465 (page 95)
    the Point of Contad B; the Angle F B C will be *equal * thiu to that in the alternate Segment of the Circle; but the Angle F B C is equal to the Angle D. Therefore the Angle in the Segment BAC will be equal to the Angle D. Therefore the Segment B A C is cut off from the given Circle ABC, containing an Angle equal to the given Right-lined Angle D; which was to he done. PROPOSITION XXXV. Theorem. ff two Right Lines in a Circle mutually cut each ether, the Rectangle contained under the Seg¬ ments of the one, is equal to the Reft angle un¬ der the Segments of the other. T N the Circle A B C D, let two Right Lines mutu- -*• ally cut each other in the Point E. I fay, the Redangle contained under A E, and E C, is equal to the Redangle contained under D E, E B. If A C and D B pafs through the Centre, fo that E be the Centre of the Circle A B C D; it is manifeft, fmee AE, EC, D E, EB, are equal; that the Redr- angle under AE, EQ is equal to the Redangle un- i der DE, EB. But if A C, D B, do not pafs through the Centre, | affume the Centre of the Circle F; from which draw i F G, F H, perpendicular to the Right Lines A C, D B; i and join F B, F C, F E. Then, becaufe the Right Line G F, drawn through ) the Centre, cuts the Right Line A C, not drawn thro’ i the Centre, at Right Angles, it will alfo bifed * the # 4 ofihih ifame. Wherefore AG is equal to GC: And be- j caufe the Right Line A C is cut into two equal Parts in the Point G, and into two unequal Parts in E, the i Redangle under A E, EC, together with the Square of EG, is f equal to the Square of GC. And if -f §,z< )the common Square of G F be added, then the Red¬ angle under AE, EC, together with the Squares of J E G, G F, is equal to the Squares of C G, G F. But .the Square of F E is f equal to the Squares of E G, 1G F, and the Square of F Q equal £ to the Squares j 47. r, of ■