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 47
... ABCD be a parallelogram : then will AB be equal to DC , and AD to BC . D For , draw the diagonal BD . Then , because AB and DC are parallel , the A angle DBA is equal to its alternate angle BDC ( P. XX . , C. 2 ) : and , because AD and ...
... ABCD be a parallelogram : then will AB be equal to DC , and AD to BC . D For , draw the diagonal BD . Then , because AB and DC are parallel , the A angle DBA is equal to its alternate angle BDC ( P. XX . , C. 2 ) : and , because AD and ...
Page 48
... ABCD , let AB be equal to DC , and AD to BC : then will it be a parallelogram . Then , the A Draw the diagonal DB . triangles ADB and CBD , will have : D the sides of the one equal to the sides of the other , each to each ; and ...
... ABCD , let AB be equal to DC , and AD to BC : then will it be a parallelogram . Then , the A Draw the diagonal DB . triangles ADB and CBD , will have : D the sides of the one equal to the sides of the other , each to each ; and ...
Page 49
... ABCD be a parallelogram , and B AC , BD , its diagonals : then will AE be equal to EC , and BE to ED . For , the triangles BEC and AED , have the angles EBC and ADE equal ( P. XX . , C. 2 ) , the angles ECB and DAE equal , and the ...
... ABCD be a parallelogram , and B AC , BD , its diagonals : then will AE be equal to EC , and BE to ED . For , the triangles BEC and AED , have the angles EBC and ADE equal ( P. XX . , C. 2 ) , the angles ECB and DAE equal , and the ...
Page 55
... A B C D ; whence , : B D A C = • If we multiply both terms of the first member by . m , and both terms of the second member by n , we shall have , mB mA = nD n C ; whence , mA : mB nC : nD ; which was to be proved . PROPOSITION IX ...
... A B C D ; whence , : B D A C = • If we multiply both terms of the first member by . m , and both terms of the second member by n , we shall have , mB mA = nD n C ; whence , mA : mB nC : nD ; which was to be proved . PROPOSITION IX ...
Page 79
... ABCD , are together equal , to two right angles ; for the angle DAB is measured by half the arc DCB , the angle DCB by half the arc B DAB hence , the two angles , taken together , are mea sured by half the circumference : hence , their ...
... ABCD , are together equal , to two right angles ; for the angle DAB is measured by half the arc DCB , the angle DCB by half the arc B DAB hence , the two angles , taken together , are mea sured by half the circumference : hence , their ...
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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