Elements of Geometry and Trigonometry |
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Page 116
... similar . But , EDF is equal to BGH : hence it is also similar to BAC ; which was to be proved . PROPOSITION XXI . THEOREM . Triangles which have their sides parallel , each to each , or perpendicular , each to each , are similar . AB ...
... similar . But , EDF is equal to BGH : hence it is also similar to BAC ; which was to be proved . PROPOSITION XXI . THEOREM . Triangles which have their sides parallel , each to each , or perpendicular , each to each , are similar . AB ...
Page 117
... similar ( P. XVIII . ) ; proved . F which was to be 2o . Let the triangles ABC and DEF have the side AB perpendicular to DE , BC to EF , and CA to FD then will they be similar . For , prolong the sides of the tri- angle DEF till they ...
... similar ( P. XVIII . ) ; proved . F which was to be 2o . Let the triangles ABC and DEF have the side AB perpendicular to DE , BC to EF , and CA to FD then will they be similar . For , prolong the sides of the tri- angle DEF till they ...
Page 118
... similar ( P. XXI . ) , we and A have , AI AF : :: DI : BF ; D R KL and , the triangles AIK and AFG , B G being similar , we have , ΔΙ : AF :: IK FG ; hence , ( B. II . , P. IV . ) , we have . DI : BF :: IK : FG . In like 118 GEOMETRY .
... similar ( P. XXI . ) , we and A have , AI AF : :: DI : BF ; D R KL and , the triangles AIK and AFG , B G being similar , we have , ΔΙ : AF :: IK FG ; hence , ( B. II . , P. IV . ) , we have . DI : BF :: IK : FG . In like 118 GEOMETRY .
Page 119
... similar to the given triangle , and to each other : 2o . Each side about the right angle will be a mean propor ... similar to ABC , and conse- quently , similar to each other . B · D The triangles ADB and ABC have the angle B common ...
... similar to the given triangle , and to each other : 2o . Each side about the right angle will be a mean propor ... similar to ABC , and conse- quently , similar to each other . B · D The triangles ADB and ABC have the angle B common ...
Page 120
... similar ( P. XVIII . , C ) . In like manner , it may be shown that the triangles ADC and ABC are similar ; and since ADB and ADC are both similar to ABC , they are similar to each other ; which was to be proved . 2o . AB will be a mean ...
... similar ( P. XVIII . , C ) . In like manner , it may be shown that the triangles ADC and ABC are similar ; and since ADB and ADC are both similar to ABC , they are similar to each other ; which was to be proved . 2o . AB will be a mean ...
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Common terms and phrases
ABCD AC² 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