Elements of Geometry and Trigonometry |
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Page 76
... similar manner , it may be shown that the fourth term cannot be less than AD : hence , it must be equal to AD ; therefore , we have , angle ACB : angle ACD :: arc AB : arc AD ; which was to be proved . Cor . 1. The intercepted arcs are ...
... similar manner , it may be shown that the fourth term cannot be less than AD : hence , it must be equal to AD ; therefore , we have , angle ACB : angle ACD :: arc AB : arc AD ; which was to be proved . Cor . 1. The intercepted arcs are ...
Page 93
... similar polygons , the parts which are similarly placed in each , are called homologous . The corresponding angles ... SIMILAR ARCS , SECTORS , or SEGMENTS , in different circles , are those which correspond to equal angles at the centre ...
... similar polygons , the parts which are similarly placed in each , are called homologous . The corresponding angles ... SIMILAR ARCS , SECTORS , or SEGMENTS , in different circles , are those which correspond to equal angles at the centre ...
Page 113
... similar . Let the triangles ABC and DEF have the angle 4 equal to the angle D , the angle B to the angle E , and the angle C to the angle F : then will they be similar . For , place the triangle DEF upon the triangle ABC , so that the ...
... similar . Let the triangles ABC and DEF have the angle 4 equal to the angle D , the angle B to the angle E , and the angle C to the angle F : then will they be similar . For , place the triangle DEF upon the triangle ABC , so that the ...
Page 114
... similar ( D. 1 ) ; which was to be proved . Cor . If two triangles have two angles in one , equal to two angles in the other , each to each , they will be similar ( B. I. , P. XXV . , C. 2 ) . PROPOSITION XIX . THEOREM . Triangles which ...
... similar ( D. 1 ) ; which was to be proved . Cor . If two triangles have two angles in one , equal to two angles in the other , each to each , they will be similar ( B. I. , P. XXV . , C. 2 ) . PROPOSITION XIX . THEOREM . Triangles which ...
Page 115
... similar . For , on BA lay off BG equal to ED ; off BH equal to EF , and draw GH . Then , : FD ; on BC lay D G because BG is equal to ED , and BH to EF , we have , BA C E. BG :: BC : BH ; hence , GH is parallel to AC ( P. XVI . ) ; and ...
... similar . For , on BA lay off BG equal to ED ; off BH equal to EF , and draw GH . Then , : FD ; on BC lay D G because BG is equal to ED , and BH to EF , we have , BA C E. BG :: BC : BH ; hence , GH is parallel to AC ( P. XVI . ) ; and ...
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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