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 22
... common angle ACE ( A. 3 ) , there remains , ACD ECB . In like manner , we find , ACD + ACE ACD + DCB ; and , taking away the common angle ACD , we have , ACE DCB . Hence , the proposition is proved . Cor . 1. If one of the angles about ...
... common angle ACE ( A. 3 ) , there remains , ACD ECB . In like manner , we find , ACD + ACE ACD + DCB ; and , taking away the common angle ACD , we have , ACE DCB . Hence , the proposition is proved . Cor . 1. If one of the angles about ...
Page 23
... common , they coin- cide throughout their whole extent , and form one and the same line . Let A and B be be two points common to two lines : then the lines coincide throughout . E B Between A and B they must coincide ( A. 11 ) . Suppose ...
... common , they coin- cide throughout their whole extent , and form one and the same line . Let A and B be be two points common to two lines : then the lines coincide throughout . E B Between A and B they must coincide ( A. 11 ) . Suppose ...
Page 24
... common angle DCA , there re- mains DCB = DCE , which is impossible , since a part can not be equal to the whole ( A. 8 ) . Hence , CB must be the prolongation of AC ; which was to be proved . PROPOSITION V. THEOREM . If two triangles ...
... common angle DCA , there re- mains DCB = DCE , which is impossible , since a part can not be equal to the whole ( A. 8 ) . Hence , CB must be the prolongation of AC ; which was to be proved . PROPOSITION V. THEOREM . If two triangles ...
Page 29
... common part AB , there remains ( A. 5 ) , 2 ° . When G is on BC . BC > EF . In this case , it is obvious that GC is less than BC ; or since GC EF , we have , = BC > EF . B CE 3o . When G is within the triangle ABC . From Proposition ...
... common part AB , there remains ( A. 5 ) , 2 ° . When G is on BC . BC > EF . In this case , it is obvious that GC is less than BC ; or since GC EF , we have , = BC > EF . B CE 3o . When G is within the triangle ABC . From Proposition ...
Page 31
... we have the side BD equal to AC , by construction , the side BC common , and the included angle ACB equal to the included angle DBC , by hypothesis : hence , the two triangles are equal in all respects ( P. V. ) . But this BOOK I. 31.
... we have the side BD equal to AC , by construction , the side BC common , and the included angle ACB equal to the included angle DBC , by hypothesis : hence , the two triangles are equal in all respects ( P. V. ) . But this BOOK I. 31.
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AB² ABCD AC² adjacent angles altitude angles is equal apothem base and altitude bisects centre chord circle circumference circumscribed cone consequently convex surface corresponding Cosine Cotang cylinder denote diagonals diameter distance divided draw drawn edges equally distant equilateral feet find the area formula frustum given angle given line given point greater hence homologous hypothenuse included angle interior angles intersection less Let ABC logarithm lower base mantissa mean proportional measured by half number of sides parallel parallelogram parallelopipedon perimeter perpendicular plane angles plane MN polyedral angle polyedron prism PROPOSITION proved pyramid quadrant radii radius rectangle regular polygon right angles right-angled triangle Scholium secant segment side BC similar sine slant height solution sphere spherical angle spherical polygon spherical triangle square Tang tangent THEOREM triangle ABC triangular prisms triedral angle upper base vertex vertices whence