Modern Methods in Elementary GeometryMacmillan, 1868 - 112 σελίδες |
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Αποτελέσματα 1 - 5 από τα 66.
Σελίδα 5
... drawn from A to 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 . Therefore AE + ...
... drawn from A to 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 . Therefore AE + ...
Σελίδα 6
... drawn B- FE D 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 B- FE D 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. F A B H Then = EFB FHL . 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 ...
... Draw EFHG cutting AB , KL , and CD , in F , H , and G. F A B H Then = EFB FHL . 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 ...
Σελίδα 12
... drawn from A to BC . Then since AB = AC , these lines B are equally remote from the perpen- dicular . A Therefore and BD = DC , C = B. I. 6. Cor . 1 . = COR . 1. Also BAD CAD . Hence in an isosceles tri angle the perpendicular bisects ...
... drawn from A to BC . Then since AB = AC , these lines B are equally remote from the perpen- dicular . A Therefore and BD = DC , C = B. I. 6. Cor . 1 . = COR . 1. Also BAD CAD . Hence in an isosceles tri angle the perpendicular bisects ...
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Algebra alternate angle altitude ARITHMETIC base BC² BEGINNERS bisect BROOKE FOSS WESTCOTT Cambridge centre chord Christ's College circle touching circumference Clifton College cloth coincide congruent CONIC SECTIONS construct contains convex angle Crown 8vo describe a circle diagonals diameter dicular distance divide Edward Thring ELEMENTARY TREATISE equal angles equally distant equiangular equilateral triangle Examination Papers Exercises exterior angle Fcap Fellow of St GEOMETRY given circle given line given point given square given straight line greater Greek Text Head Master Hence inscribe intersect John's College Join JOSEPH WOLSTENHOLME Jugurtha late Fellow Latin Let ABCD line drawn locus Mathematical Lecturer measure meet middle point MIXED MATHEMATICS numerous parallel to BC parallelogram perpen perpendicular polygon problems quadrilateral figure radius rectangle right angles Second Edition Shew similar triangles student tangent THEOREM VII Third Edition trapezium triangle ABC vertex
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