Elements of Geometry and Trigonometry: From the Works of A. M. Legendre |
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Page 65
... circumference may be made to pass , and but one . Let A , B , and C , be any three points , not in a straight line then may one circumference be made to pass through them , and but one . Join the points by the lines AB , BC , and bisect ...
... circumference may be made to pass , and but one . Let A , B , and C , be any three points , not in a straight line then may one circumference be made to pass through them , and but one . Join the points by the lines AB , BC , and bisect ...
Page 66
... circumference may be made to pass through these points , and but one ; which was to be proved . Cor . Two circumferences cannot intersect in more than two points ; for , if they could intersect in three points , there would be two ...
... circumference may be made to pass through these points , and but one ; which was to be proved . Cor . Two circumferences cannot intersect in more than two points ; for , if they could intersect in three points , there would be two ...
Page 68
... circumference at the B A E D Ο point A ; it is , therefore , tangent to it at that point ( D. 11 ) ; which was to be proved . 2o . Let BD be tangent to the circle at A : then will it be perpendicular to CA. For , let E be any point of ...
... circumference at the B A E D Ο point A ; it is , therefore , tangent to it at that point ( D. 11 ) ; which was to be proved . 2o . Let BD be tangent to the circle at A : then will it be perpendicular to CA. For , let E be any point of ...
Page 69
... circumference . There may be three cases : both parallels may be secants ; one may be a secant and the other a tangent ; or , both may be tangents . 1o . Let the secants AB and DE be parallel : then will the intercepted arcs MN and PQ ...
... circumference . There may be three cases : both parallels may be secants ; one may be a secant and the other a tangent ; or , both may be tangents . 1o . Let the secants AB and DE be parallel : then will the intercepted arcs MN and PQ ...
Page 70
... circumferences intersect each other , the points of in- tersection will be in a perpendicular to the line joining their ... circumference whose centre is D , they are equally dis- tant from D : hence , CD is perpendicular to AB at its ...
... circumferences intersect each other , the points of in- tersection will be in a perpendicular to the line joining their ... circumference whose centre is D , they are equally dis- tant from D : hence , CD is perpendicular to AB at its ...
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Common terms and phrases
AB² ABCD altitude apothem Applying logarithms centre chord circle circumference cone consequently convex surface cosec Cosine Cotang cylinder demonstrated in Book denote diameter distance divided draw edges Equation feet find the area Find the logarithmic following RULE frustum given angle greater hence homologous hypothenuse included angle inscribed intersection less Let ABC linear units log cot log sin lower base lune mantissa multiplied number of sides opposite parallel parallelogram parallelopipedon perpendicular plane MN polar triangle polyedral angle polyedron principle demonstrated prism proportional PROPOSITION proved pyramid quadrant radii radius rectangle regular polygon right angles right-angled triangle Scholium segment similar six right slant height solution sphere spherical angle spherical excess spherical polygon spherical triangle square straight line subtracting Tang tangent THEOREM triangle ABC triangular prism upper base vertex volume whence write the following