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
... point C lies within the triangle ADF , the sum of the lines AD and DF will be greater than the sum of the lines AC ... middle point : 10. Any point of the perpendicular will be equally distant 1o . from the extremities of the line : 2o ...
... point C lies within the triangle ADF , the sum of the lines AD and DF will be greater than the sum of the lines AC ... middle point : 10. Any point of the perpendicular will be equally distant 1o . from the extremities of the line : 2o ...
Page 36
... points E and F equally distant from A and B , it will be perpen- dicular to the line AB at its middle point . PROPOSITION XVII . THEOREM . If two right - angled triangles have the hypothenuse and of the one equal to the hypothenuse and ...
... points E and F equally distant from A and B , it will be perpen- dicular to the line AB at its middle point . PROPOSITION XVII . THEOREM . If two right - angled triangles have the hypothenuse and of the one equal to the hypothenuse and ...
Page 38
... 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 .
... 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