. The Elements of Euclid : viz. the first six books, together with the eleventh and twelfth : the errors, by which Theon, or others, have long ago vitiated these books, are corrected, and some of Euclid's demonstrations are restored : also, the book of Euclid's Data, in like manner corrected. from their centres are equal; therefore BK, KC are equal toEL, LF; and the base BC is equal to the base EF; thereforethe angle BKC is equal ^ to the angle ELF: but equal angles b upon equal <= circumferences, when they are at the ; therefore the circumference BGC is equal to the
. The Elements of Euclid : viz. the first six books, together with the eleventh and twelfth : the errors, by which Theon, or others, have long ago vitiated these books, are corrected, and some of Euclid's demonstrations are restored : also, the book of Euclid's Data, in like manner corrected. from their centres are equal; therefore BK, KC are equal toEL, LF; and the base BC is equal to the base EF; thereforethe angle BKC is equal ^ to the angle ELF: but equal angles b upon equal <= circumferences, when they are at the ; therefore the circumference BGC is equal to the circum-ference EHF. But the whole circle ABC is equal to the wholeEDF; the remaining part, therefore, of the circumference, , is equal to the remaining part EDF. Therefore, in equalcircles, &:c. Q. E. D. PROP. XXIX. THEOR. IN equal circles, equal circumferences are subtendedby equal straight lines. Let ABC, DEF be equal circles, and let the circumferencesBGC, EHF also be equal; and join BC, EF: the straight lineBC is equal to the straight line EF. 92 THE ELEMENTS Book III. Take » K, L, the centres of the circles, and join BK, KC,v^.iV—^ EL, LF: and because the circumference BGC is equal to thea 1. 3. A D. G H circumference EHF, the angle BKC is equal ^ to the angleELF: and because the circles ABC, DEF are equal, the straightlines from their centres are equal: therefore BK, KC are equalto EL, LF, and they contain equal angles: therefore the base c BC is equal <= to the base EF. Therefore, in equal circles, E. D. PROP. XXX. PROB. a 10. 1. TO bisect a given circumference, that is, to dhideit into two equal parts. Let ADB be the given circumference ; it is required to bisect it. Join AB, and bisect » it in C; from the point C draw CD atright angles to AB, and join AD, DB: the circumference ADBis bisected in the point D. Because AC is equal to CB, and CD common to the trianglesACD, BCD, the two sides AC, CDare equal to the two BC, CD; andthe angle A
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Keywords: ., bookauthoreuclid, bookcentury1800, booksubje, booksubjectgeometry
