Elements of Geometry and Trigonometry: From the Works of A. M. Legendre |
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Page 26
... hence , the triangles coincide throughout , and are therefore equal in all their parts ( I. , D. 14 ) ; which was to be proved . · PROPOSITION VII . THEOREM . The sum of any two sides of a triangle is greater than the third side . Let ...
... hence , the triangles coincide throughout , and are therefore equal in all their parts ( I. , D. 14 ) ; which was to be proved . · PROPOSITION VII . THEOREM . The sum of any two sides of a triangle is greater than the third side . Let ...
Page 31
... hence , the triangles BAD , and DAC , have the three sides of the one equal to those of the other , each to each ; therefore , by the last Proposition , the angle B is equal to the angle C ;. which was to be proved . B D Cor . 1. An ...
... hence , the triangles BAD , and DAC , have the three sides of the one equal to those of the other , each to each ; therefore , by the last Proposition , the angle B is equal to the angle C ;. which was to be proved . B D Cor . 1. An ...
Page 32
... hence , the hypothesis that AB and AC are unequal , is false . must , therefore , be equal ; which was to be proved . Cor . An equiangular triangle is equilateral . PROPOSITION XIII . THEOREM . In any triangle , the greater side is ...
... hence , the hypothesis that AB and AC are unequal , is false . must , therefore , be equal ; which was to be proved . Cor . An equiangular triangle is equilateral . PROPOSITION XIII . THEOREM . In any triangle , the greater side is ...
Page 33
... hence , the angles ACB and FCB are equal ( P. V. ) But ACB is , by a hypothesis , a right angle : hence , FCB must also be a right angle , and consequently , the line ACF must be a straight line ( P. IV . ) . But this is impos- sible ...
... hence , the angles ACB and FCB are equal ( P. V. ) But ACB is , by a hypothesis , a right angle : hence , FCB must also be a right angle , and consequently , the line ACF must be a straight line ( P. IV . ) . But this is impos- sible ...
Page 34
... hence , FC is equal to AC ( P. V. ) . But , AF is shorter than ACF ( A. 12 ) : hence , AB , the half of AF , is shorter than AC , the half of ACF ; which was to be proved . 2o . In the triangles ABC and ABE , we have the side BC equal ...
... hence , FC is equal to AC ( P. V. ) . But , AF is shorter than ACF ( A. 12 ) : hence , AB , the half of AF , is shorter than AC , the half of ACF ; which was to be proved . 2o . In the triangles ABC and ABE , we have the side BC equal ...
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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