Elements of Geometry and Trigonometry |
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Page 34
... equally distant from the foot of the perpendicular , will be equal : 3 ° . Of two oblique lines that meet the given line at points unequally distant from the foot of the perpendicular , the one which meets it at the greater distance ...
... equally distant from the foot of the perpendicular , will be equal : 3 ° . Of two oblique lines that meet the given line at points unequally distant from the foot of the perpendicular , the one which meets it at the greater distance ...
Page 35
... equally dis- tant from A and B ; and any point without EF , will be unequally distant from A and B. A 1o . From any point of EF , as D , draw the lines DA and DB . Then will DA and DB be equal ( P. XV . ) : hence , Dis B equally distant ...
... equally dis- tant from A and B ; and any point without EF , will be unequally distant from A and B. A 1o . From any point of EF , as D , draw the lines DA and DB . Then will DA and DB be equal ( P. XV . ) : hence , Dis B equally distant ...
Page 36
... 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 ; which was to be proved . A D B Cor . If a straight line EF have two of its points E and F equally distant from A and B , it ...
... 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 ; which was to be proved . A D B Cor . If a straight line EF have two of its points E and F equally distant from A and B , it ...
Page 42
... equally distant . Let AB and CD be parallel : then will they be every . where equally distant . CH A GD B From any two points of AB , as F and E , draw FI and EG perpendicular to CD ; they will also be Ferpendicular to AB ( P. XX . , C ...
... equally distant . Let AB and CD be parallel : then will they be every . where equally distant . CH A GD B From any two points of AB , as F and E , draw FI and EG perpendicular to CD ; they will also be Ferpendicular to AB ( P. XX . , C ...
Page 43
... equal to GFE , GFH equal to FGE and the side FG common ; they are , therefore , equal in all their parts ( P. VI . ) hence , FH is equal to EG ; and consequently , AB and CD are everywhere equally distant ; which was to be proved ...
... equal to GFE , GFH equal to FGE and the side FG common ; they are , therefore , equal in all their parts ( P. VI . ) hence , FH is equal to EG ; and consequently , AB and CD are everywhere equally distant ; which was to be proved ...
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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