Elements of Geometry and Trigonometry from the Works of A.M. Legendre: Adapted to the Course of Mathematical Instruction in the United States |
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Page 27
... remaining sides of the triangle . A Let be any point within the triangle BAC , and let the lines OB , OC , be drawn to the extremities of any side , as BC : then will the sum of BO and OC be less than the sum of the sides BA and AC ...
... remaining sides of the triangle . A Let be any point within the triangle BAC , and let the lines OB , OC , be drawn to the extremities of any side , as BC : then will the sum of BO and OC be less than the sum of the sides BA and AC ...
Page 61
... remaining common ; E then will they coincide ; otherwise there would be some points in either one or the other of the curves unequally distant from the centre ; which is impossible ( D. 1 ) : hence , AB divides the circle , and also its ...
... remaining common ; E then will they coincide ; otherwise there would be some points in either one or the other of the curves unequally distant from the centre ; which is impossible ( D. 1 ) : hence , AB divides the circle , and also its ...
Page 110
... remaining terms are proportional ( B. II . , P. IV . ) , hence , AD : DB :: AE : EC ; which was to be proved . Cor . 1. We have , by composition ( B. II . , P. VI . ) , AD + DB : AD :: AE + EC : AE ; or , AB AD :: AC : AE ; and 110 ...
... remaining terms are proportional ( B. II . , P. IV . ) , hence , AD : DB :: AE : EC ; which was to be proved . Cor . 1. We have , by composition ( B. II . , P. VI . ) , AD + DB : AD :: AE + EC : AE ; or , AB AD :: AC : AE ; and 110 ...
Page 127
... remaining segment of the second is to the remaining segment of the first . For , draw CD and BA . Then will the angles ODB and OAC be equal , because each is measured by half of the arc CB ( B. III . , P. XVIII . ) . The angles OBD and ...
... remaining segment of the second is to the remaining segment of the first . For , draw CD and BA . Then will the angles ODB and OAC be equal , because each is measured by half of the arc CB ( B. III . , P. XVIII . ) . The angles OBD and ...
Page 203
... remaining pyramid may be regarded as having the triangle Ff for its base , and the point g for its vertex . From g , draw gK parallel to ƒF , and draw also KH and Kf . Then will the pyramids K - FfH and g - FfII , be equal ; for they ...
... remaining pyramid may be regarded as having the triangle Ff for its base , and the point g for its vertex . From g , draw gK parallel to ƒF , and draw also KH and Kf . Then will the pyramids K - FfH and g - FfII , be equal ; for they ...
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Common terms and phrases
AB² AC² altitude angle ACB apothem axis base and altitude base multiplied BC² bisect centre chord circumference coincide cone consequently convex surface corresponding cosec cosine Cotang cylinder denote diameter distance divided draw drawn edges equal bases equal in volume equal to AC equal to half equally distant Formula frustum given angle given line greater hence homologous hypothenuse included angle intersection less Let ABC logarithm lower base mantissa mean proportional measured by half number of sides opposite parallelogram perimeter perpendicular plane MN polyedral angle polyedron prism PROPOSITION XI proved pyramid quadrant radii radius rectangle regular polygons right-angled triangle Scholium segment semi-circumference side BC similar sine slant height sphere spherical angle spherical excess spherical polygon spherical triangle straight line tangent THEOREM triangle ABC triangular prisms triedral angle upper base vertex vertices whence