Euclidian GeometryMacmillan, 1874 - 349 sider |
Inni boken
Side 36
... possible , be produced to meet towards Q , S in the point X. Then GHX is a ; and the exterior △ PGH is > the interior and opposite L GHX ; but it is also equal to it , which is impossible . ( I. 13 ) .. PQ , RS do not meet when ...
... possible , be produced to meet towards Q , S in the point X. Then GHX is a ; and the exterior △ PGH is > the interior and opposite L GHX ; but it is also equal to it , which is impossible . ( I. 13 ) .. PQ , RS do not meet when ...
Side 38
... possible , at F in FG draw another line XFY , making L XFG = L FGD . Then XFY is || to CD . ( I. 20 ) .. through F two straight lines have been drawn || to CD , which is impossible .. .. LAFG is the alternate FGD . = ( Lemma . ) Hence ...
... possible , at F in FG draw another line XFY , making L XFG = L FGD . Then XFY is || to CD . ( I. 20 ) .. through F two straight lines have been drawn || to CD , which is impossible .. .. LAFG is the alternate FGD . = ( Lemma . ) Hence ...
Side 64
... possible positions of that point . PROBLEM ( c ) . Find the locus of a point equidistant from two given points . Let A , B be the two given . points . Bisect AB in P , then Pis one point in the locus . Let be any other point in the ...
... possible positions of that point . PROBLEM ( c ) . Find the locus of a point equidistant from two given points . Let A , B be the two given . points . Bisect AB in P , then Pis one point in the locus . Let be any other point in the ...
Side 76
... possible upon the same base BC and on the same side of it let there be two as ABC , DBC which have their conterminous sides equal , viz . AB = DB__and AC DC . = Ist . Let the vertex of each a fall without the other a . B A D Join AD ...
... possible upon the same base BC and on the same side of it let there be two as ABC , DBC which have their conterminous sides equal , viz . AB = DB__and AC DC . = Ist . Let the vertex of each a fall without the other a . B A D Join AD ...
Side 95
... possible , let it fall without the as AQB . Join C with A and B ; R A and draw a straight line CRQ cutting the circle in R , and the straight line AB in Q. Then . CA = CB ; .. L CAB : = L CBA ; but L CQB is > CAB ; .. 4 CQB is > △ CBA ...
... possible , let it fall without the as AQB . Join C with A and B ; R A and draw a straight line CRQ cutting the circle in R , and the straight line AB in Q. Then . CA = CB ; .. L CAB : = L CBA ; but L CQB is > CAB ; .. 4 CQB is > △ CBA ...
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Algebra Arithmetic base BEGINNERS Cambridge centre chord Christ's College circumference cloth Conic Sections Crown 8vo Describe a circle diagonals diameter divided ELEMENTARY TREATISE English equiangular equilateral Euclid Examples Extra fcap GEOMETRY given angle given circle given point given straight line Grammar greater H Let Hence inscribed intersecting isosceles triangle John's College Latin Let ABC line bisecting locus magnitudes Mathematical meet Owens College parallel parallelogram perimeter perpendicular plane PROBLEM produced Professor proportional PROPOSITION radius ratio rect rectangle rectangle contained rectilineal figure regular polygon render respectively revised rhombus right angles Schools Second Edition segment Similarly squares on AC straight line joining student tangent THEOREM TRIGONOMETRY twice rectangle twice the squares Uppingham School vertex
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Side 157 - The first of four magnitudes is said to have the same ratio to the second, which the third has to the fourth, when any equimultiples whatsoever of the first and third being taken, and any equimultiples whatsoever of the second and fourth ; if the multiple of the first be less than that of the second, the multiple of the third is also less than that of the fourth...
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