Elements of Geometry and Trigonometry |
From inside the book
Results 1-5 of 48
Page 23
... until the assumed hypothesis is shown to be false . Its contradictory is thus proved to be true . This method of demonstration is often used in Geometry . PROPOSITION IV . THEOREM . If a straight line meet BOOK I. 23.
... until the assumed hypothesis is shown to be false . Its contradictory is thus proved to be true . This method of demonstration is often used in Geometry . PROPOSITION IV . THEOREM . If a straight line meet BOOK I. 23.
Page 27
... , we have , BD + DC < BA + AC . But it was shown that BO + OC is less than BD + - DC ; still more , then , is BO + OC less than BA + AC ; which was to be proved . PROPOSITION IX . THEOREM . If two triangles have two BOOK 1 . 27.
... , we have , BD + DC < BA + AC . But it was shown that BO + OC is less than BD + - DC ; still more , then , is BO + OC less than BA + AC ; which was to be proved . PROPOSITION IX . THEOREM . If two triangles have two BOOK 1 . 27.
Page 35
... shown , as in the first case , that AD is equal to DF . Then , because the 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 and CF ( P. VIII . ) : hence , AD , the half ...
... shown , as in the first case , that AD is equal to DF . Then , because the 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 and CF ( P. VIII . ) : hence , AD , the half ...
Page 36
... shown , DA and DB will be equal ; but IB is less than the sum of ID and DB ( P. VII . ) ; and because the sum of ID and DB is equal to the sum of ID and DA , or IA , we have IB less than IA : hence , I is unequally distant from A and B ...
... shown , DA and DB will be equal ; but IB is less than the sum of ID and DB ( P. VII . ) ; and because the sum of ID and DB is equal to the sum of ID and DA , or IA , we have IB less than IA : hence , I is unequally distant from A and B ...
Page 39
... shown , AB and CD are parallel . Cor . 2. If two straight lines are cut by a third , making the opposite exterior and interior angles equal , the two straight lines will be parallel . Let the angles EGB and GHD be equal : Now EGB and ...
... shown , AB and CD are parallel . Cor . 2. If two straight lines are cut by a third , making the opposite exterior and interior angles equal , the two straight lines will be parallel . Let the angles EGB and GHD be equal : Now EGB and ...
Other editions - View all
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