Geometry without axioms; or the first book of Euclid's elements. With alterations and familiar notes and an intercalary book in which the straight ... of the sphere . To which is added an appendix - Softcover

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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. 1833 Excerpt: ... to A, B, C, respectively. t1. 3. In a straight line DE, maket FG equal to any one of the straight lines as A; FH equal to another as B; and GI equal to Jinterc.13. C. About the centre F, with the radius FH, describej a circle; interc.13. ana let cut E n poult N. About the centre G, with the Cor. 5. radius GI, describe another circle; and because FH and GI are tHyp. togetherf greater than FG, this last circle will cut GD in a point (as M) nearer to D than is the point N. For if M and N should coincide, FH and GI must be together equal to FG; and if M should lie on the other side of N, they must be together less than I Hyp. FG; neither of which is possible, for they arej greater. Also Hyp. because FN (which is equal to FH) is less than the sum of FG and GI, the point N will fall on the side of I which is towards D; t Hyp. and because GM (which is equal to GI) isf less than the sum of GF and FH) the point M will fall on the side of H which is towards E. Wherefore the circumference of each of the circles will pass through that of the other; and in passing from the outside of it to the inside, and from the inside to the outside again, JI. 7. Cor. will necessarily cut it in two pointsj. Let one of these be K. Join KF, KG. The triangle FKG has its sides equal to the three straight lines A, B, C. interc.13. Because the point F is the centre of the circle HKL, FK is CContt "V to FH. But FH isf equal to the straight line B; thereJinterc. 1. fore FK ist equal to B. Again, because G is the centre of the Cor. 1. circle IKL, GK is equal to GI. But GI is equal to the straight line C; therefore GK is equal to C. Also FG is equal to A; therefore the three straight lines FG, FK, GK (which are the sides of the triangle FKG) are respectively equal to the three straight l...

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