Elements of Geometry and Trigonometry |
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Page 31
... middle point D of the base equal to AC , by hypothesis , AD common , and BD equal to DC , by construction : hence , the triangles BAD , and DAC , have the three sides of the one equal to those of the other , each to each ; therefore ...
... middle point D of the base equal to AC , by hypothesis , AD common , and BD equal to DC , by construction : hence , the triangles BAD , and DAC , have the three sides of the one equal to those of the other , each to each ; therefore ...
Page 35
... middle point : 10. Any point of the perpendicular will be equally distant 1o . from the extremities of the line : 2o . Any point , without the perpendicular , will be unequally distant from the extremities . Let AB be a given straight ...
... middle point : 10. Any point of the perpendicular will be equally distant 1o . from the extremities of the line : 2o . Any point , without the perpendicular , will be unequally distant from the extremities . Let AB be a given straight ...
Page 36
... middle point . PROPOSITION XVII . THEOREM . If two right - angled triangles have the hypothenuse and of the one equal to the hypothenuse and a side of the other , each to each , the triangles will be equal in all their parts . Let the ...
... middle point . PROPOSITION XVII . THEOREM . If two right - angled triangles have the hypothenuse and of the one equal to the hypothenuse and a side of the other , each to each , the triangles will be equal in all their parts . Let the ...
Page 38
... KC and HD be parallel . Through G , the middle point of AB , draw GF perpendicular to KC , and prolong it to E. The sum of the angles GBE and GBD is equal to two right EB H -D G K AF 1 angles ( P. I. ) ; the sum of the 38 GEOMETRY .
... KC and HD be parallel . Through G , the middle point of AB , draw GF perpendicular to KC , and prolong it to E. The sum of the angles GBE and GBD is equal to two right EB H -D G K AF 1 angles ( P. I. ) ; the sum of the 38 GEOMETRY .
Page 64
... hypothenuse CA equal to CB , and the side CD A B common ; the triangles are , therefore , equal in all their parts hence , AD is equal to DB . Again , because CG . is perpendicular to AB , at its middle point , 64 GEOMETRY .
... hypothenuse CA equal to CB , and the side CD A B common ; the triangles are , therefore , equal in all their parts hence , AD is equal to DB . Again , because CG . is perpendicular to AB , at its middle point , 64 GEOMETRY .
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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