Elements of Geometry and Trigonometry |
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Page 47
... ABCD be a parallelogram : then will AB be equal to DC , and AD to BC . For , draw the diagonal BD . Then , because AB and DC are parallel , the angle DBA is equal to its alternate A angle BDC ( P. XX . , C. 2 ) : and , because AD and BC ...
... ABCD be a parallelogram : then will AB be equal to DC , and AD to BC . For , draw the diagonal BD . Then , because AB and DC are parallel , the angle DBA is equal to its alternate A angle BDC ( P. XX . , C. 2 ) : and , because AD and BC ...
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 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 DAB hence , the two angles , taken together , are mea- sured by half the circumference : hence , their sum ...
... 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 DAB hence , the two angles , taken together , are mea- sured by half the circumference : hence , their sum ...
Page 90
... ABCD ; join the point A with the points of intersection D and B : then will both AD and AB be tangent to the given circle , and there will be two solutions . For , the angles ABC and ADC are right angles ( P. XVIII . , C. 2 ) : hence ...
... ABCD ; join the point A with the points of intersection D and B : then will both AD and AB be tangent to the given circle , and there will be two solutions . For , the angles ABC and ADC are right angles ( P. XVIII . , C. 2 ) : hence ...
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Common terms and phrases
ABCD ACĀ² adjacent angles altitude angle ACB apothem Applying logarithms base and altitude bisect centre chord circle circumference circumscribed coincide cone consequently convex surface corresponding cosec cosine Cotang cylinder denote diagonals diameter distance divided draw drawn edges equally distant feet find the area Formula frustum given angle given line given point greater hence homologous hypothenuse included angle interior angles intersection less Let ABC log sin logarithm lower base mantissa mean proportional measured by half middle point number of sides opposite parallel parallelogram parallelopipedon perimeter perpendicular plane MN polyedral angle polyedron prism PROBLEM PROPOSITION proved pyramid quadrant radii radius rectangle regular polygons right angles right-angled triangle Scholium secant segment side BC similar sine slant height sphere spherical polygon spherical triangle square Tang tangent THEOREM triangle ABC triangular prisms triedral angle upper base vertex vertices whence