Plane and Solid Geometry: To which is Added Plane and Spherical Trigonometry and Mensuration. Accompanied with All the Necessary Logarithmic and Trigonometric Tables |
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Page 14
... consequently , BFC + CFA = CFA + AFD ( A. I. ) : from each taking CFA , we have BFC equal to its opposite angle AFD . In a similar manner we have CFA = DFB . Cor . I. If either of the four angles , formed by the intersection of two ...
... consequently , BFC + CFA = CFA + AFD ( A. I. ) : from each taking CFA , we have BFC equal to its opposite angle AFD . In a similar manner we have CFA = DFB . Cor . I. If either of the four angles , formed by the intersection of two ...
Page 15
... consequently supplementary with its equal angle BAC . E THEOREM III . Two angles having their corresponding sides ... consequently the angle BAC is equal to KGC . And since AC and DF are parallel they have the same direc- tion in ...
... consequently supplementary with its equal angle BAC . E THEOREM III . Two angles having their corresponding sides ... consequently the angle BAC is equal to KGC . And since AC and DF are parallel they have the same direc- tion in ...
Page 16
... consequently , the angle KGC is equal to EDF ; therefore ( A. I. ) , the angle BAC is equal to EDF . Secondly . When the sides AB and AC are respectively in opposite directions to DE and DF . As before , we have the angle BAC equal to ...
... consequently , the angle KGC is equal to EDF ; therefore ( A. I. ) , the angle BAC is equal to EDF . Secondly . When the sides AB and AC are respectively in opposite directions to DE and DF . As before , we have the angle BAC equal to ...
Page 19
... consequently ( A. IV . ) , AD + DB < AD + DE + EB , that is , AD + DB < AE + EB . Again , we have AE < AC + CE , consequently , AE + EB < AC + CE + EB , that is , AE + EBAC + CB . Comparing these conditions , we have AD + DB < AC + CB ...
... consequently ( A. IV . ) , AD + DB < AD + DE + EB , that is , AD + DB < AE + EB . Again , we have AE < AC + CE , consequently , AE + EB < AC + CE + EB , that is , AE + EBAC + CB . Comparing these conditions , we have AD + DB < AC + CB ...
Page 23
... Consequently the perpendiculars DE and DF will coincide ( T. XI . ) , and are therefore equal . Secondly . Suppose G to be a point without the bisecting line . If we draw the perpendiculars GF and GH , one of these must cut the ...
... Consequently the perpendiculars DE and DF will coincide ( T. XI . ) , and are therefore equal . Secondly . Suppose G to be a point without the bisecting line . If we draw the perpendiculars GF and GH , one of these must cut the ...
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Common terms and phrases
a+b+c ACē altitude angles equal apothem bisect centre chord circ circumference cone consequently corresponding cosec Cosine Cotang cylinder decimal denote described diameter dicular distance divided draw drawn equation equivalent exterior angles feet figure formed frustum give given line greater half hence homologous sides hypotenuse inscribed circle intersection logarithm measure middle point multiplied number of sides opposite parallelogram parallelopipedon pendicular perimeter perpen perpendicular plane MN polyedral angle polyedron prism PROBLEM proportion pyramid quadrant radii radius ratio rectangle regular inscribed regular polygon respectively equal right angles right-angled triangle Scholium secant sector similar similar triangles Sine slant height solid sphere spherical triangle square straight line subtract suppose surface Tang tangent THEOREM three sides triangle ABC triangular prism vertex volume