. 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. GH, HK,each to each, and they contain equal angles; wherefore thebase CP is equal to the base GK, that is, to LN. And in theisosceles triangles ABC, MXL, because the angle ABC is great-er than the angle MXL, therefore the angle MLX at the base isgreater s than the angle ACB at the base. For the same rea-
. 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. GH, HK,each to each, and they contain equal angles; wherefore thebase CP is equal to the base GK, that is, to LN. And in theisosceles triangles ABC, MXL, because the angle ABC is great-er than the angle MXL, therefore the angle MLX at the base isgreater s than the angle ACB at the base. For the same rea- g 32. >.son, because the angle GHK, or CBP, is greater than the angle R LXN, the angle XLN is greaterthan the angle BCP. Thereforethe whole angle MLN is greaterthan the whole angle ACP. Andbecause ML, LN are equal toAC, CP, each to each, but theangle MLN is greater than theangle ACP, the base MN isgreater ^ than the base MN is equal to DF ; there-fore also DF is greater than , because DE,EF are equalto AB, BP, but the base DFgreater than the base AP, theangle DEF is greater ^ than the 1- 35. l angle ABP. And ABP is equal to the two angles ABC, CBP,that is, to the two angles ABC, GHK; therefore the angle DEFis greater than the two angles ABC, GHK ; but it is also less. h 220 THE ELEMENTS b 3. Book XI. than these, which is impossible. Therefore AB is not less than Ci-y-.^ LX, and it has been proved that it is not equal to it; thereforeAB is greater than LX. a From the point X erect a XR at right angles to the plane ofthe circle LMN. And because it has been proved in all the casesthat AB is greater than LX, find a square equal to the excess ofthe square of AB above the squareof LX; and make RX equal to itsside, and join RL, RM, RN. Be-cause RX is perpendicular to theplane of the circle LMN, it is •> per-pendicular to each of the straightlines LX, MX, NX. And becauseLX is equal to MX, and XR com-mon, and at right angles to each ofthem, th
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Keywords: ., bookauthoreuclid, bookcentury1800, booksubje, booksubjectgeometry
