Elements of Geometry and Trigonometry: From the Works of A. M. Legendre |
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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 00 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 00 be less than the sum of the sides BA and AC ...
Page 61
... remaining common ; 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 ; 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 : ᎠᏴ :: AE : EC ; which was to be proved . Cor . 1. We have , by composition ( B. II . , P. VI . ) , AD + DB : AD :: AE EC : AE ; AB : AD :: AC : AE ; or , 110 GEOMETRY .
... remaining terms are proportional ( B. II . , P. IV . ) , hence , AD : ᎠᏴ :: AE : EC ; which was to be proved . Cor . 1. We have , by composition ( B. II . , P. VI . ) , AD + DB : AD :: AE EC : AE ; AB : AD :: AC : AE ; or , 110 GEOMETRY .
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 FfH for its base , and the point g for its vertex . From g , draw gK parallel to fF , and draw also KH and Kf . Then will the pyramids K - FfH and g - FfH , be equal ; for they ...
... remaining pyramid may be regarded as having the triangle FfH for its base , and the point g for its vertex . From g , draw gK parallel to fF , and draw also KH and Kf . Then will the pyramids K - FfH and g - FfH , be equal ; for they ...
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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