Modern Methods in Elementary GeometryMacmillan, 1868 - 112 σελίδες |
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Σελίδα 5
... drawn from A to B. α A B THEOREM IV . Two straight lines AC , CB are less than two straight lines AD , DB which enclose them . Produce AC to meet DB in E. E Then CB CE + EB . Therefore AC + CB < AE + EB . Again , AE < AD + DE ...
... drawn from A to B. α A B THEOREM IV . Two straight lines AC , CB are less than two straight lines AD , DB which enclose them . Produce AC to meet DB in E. E Then CB CE + EB . Therefore AC + CB < AE + EB . Again , AE < AD + DE ...
Σελίδα 6
... drawn FE D B- -C from a point A to a straight line BC ; and of the others AE which is nearer to the perpendicular is always less than one more remote AF Let the figure AFD be turned about BC so that A may fall on A ' . Then , since ADB ...
... drawn FE D B- -C from a point A to a straight line BC ; and of the others AE which is nearer to the perpendicular is always less than one more remote AF Let the figure AFD be turned about BC so that A may fall on A ' . Then , since ADB ...
Σελίδα 10
... drawn each parallel to CD , which is absurd . COR . 1. The other angles at G are equal to the corre- sponding angles at H. COR . 2. The alternate angles AGH , GHD are equal . COR . 3. The two interior angles BGH , GHD are toge- ther ...
... drawn each parallel to CD , which is absurd . COR . 1. The other angles at G are equal to the corre- sponding angles at H. COR . 2. The alternate angles AGH , GHD are equal . COR . 3. The two interior angles BGH , GHD are toge- ther ...
Σελίδα 11
... Draw EFHG cutting AB , KL , and CD , in F , H , and G. A. F B H Then EFBFHL . K L G And = FHL HGD . C D Therefore EFB = HGD . Therefore AB is parallel to CD . I. 7 . TRIANGLES . THEOREM X. If one side BC of a triangle ABC be produced ...
... Draw EFHG cutting AB , KL , and CD , in F , H , and G. A. F B H Then EFBFHL . K L G And = FHL HGD . C D Therefore EFB = HGD . Therefore AB is parallel to CD . I. 7 . TRIANGLES . THEOREM X. If one side BC of a triangle ABC be produced ...
Σελίδα 12
... drawn from A to BC . Then since AB = AC , these lines are equally remote from the perpen- dicular . A A B D Therefore and BD = DC , C = B. I. 6. Cor . 1 . = COR . 1. Also BAD CAD . Hence in an isosceles tri- angle the perpendicular ...
... drawn from A to BC . Then since AB = AC , these lines are equally remote from the perpen- dicular . A A B D Therefore and BD = DC , C = B. I. 6. Cor . 1 . = COR . 1. Also BAD CAD . Hence in an isosceles tri- angle the perpendicular ...
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A'BC Algebra alternate angle altitude angle equal arc CD ARITHMETIC BC² bisect BROOKE FOSS WESTCOTT CA² Cambridge centre chord circle touching circumference cloth coincide congruent CONIC SECTIONS contain convex angle convex polygon Crown 8vo diagonals diameter dicular divide Edward Thring ELEMENTARY TREATISE equal angles equally distant equally remote equiangular equiangular and similar equilateral triangle Examples exterior angle Fcap Find the locus GEOMETRY given angle given circle given line given point given straight line greater Hence intersect isosceles John's College late Fellow Let ABCD line drawn MATHEMATICAL measure meet middle point opposite sides parallel to CD parallelogram perpen perpendicular polygon produced quadrilateral figure radius rectangle right angles right-angled triangle Schools Second Edition segments Shew shortest line similar triangles student tangent THEOREM THEOREM VII Third Edition trapezium triangle ABC vertex