Modern Methods in Elementary GeometryMacmillan, 1868 - 112 σελίδες |
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Αποτελέσματα 1 - 5 από τα 13.
Σελίδα 8
... Parallel straight lines are such as being produced ever so far both ways do not meet . If a straight line EF cut the two straight lines AB , CD , it makes with either of them four angles which we will denote by ( 1 ) , ( 2 ) , ( 3 ) ...
... Parallel straight lines are such as being produced ever so far both ways do not meet . If a straight line EF cut the two straight lines AB , CD , it makes with either of them four angles which we will denote by ( 1 ) , ( 2 ) , ( 3 ) ...
Σελίδα 9
... CD do not meet either towards B and D , or towards C and A. COR . 1. If either of the angles at G be equal to the corresponding angle at H , AB must be parallel to CD . COR . 2. If the alternate angle AGH is equal to the alternate angle ...
... CD do not meet either towards B and D , or towards C and A. COR . 1. If either of the angles at G be equal to the corresponding angle at H , AB must be parallel to CD . COR . 2. If the alternate angle AGH is equal to the alternate angle ...
Σελίδα 10
... parallel to CD . For BGH + GHD will then be equal to BGH + EGB , and therefore EGB = GHD . THEOREM VIII . If AB be parallel to CD , then shall the angle EGB be E L A- -B M H C- D equal to the angle GHD . For if not let EGM be equal to ...
... parallel to CD . For BGH + GHD will then be equal to BGH + EGB , and therefore EGB = GHD . THEOREM VIII . If AB be parallel to CD , then shall the angle EGB be E L A- -B M H C- D equal to the angle GHD . For if not let EGM be equal to ...
Σελίδα 11
E. M. Reynolds. THEOREM IX . Straight lines AB , CD parallel to the same line KL are parallel to one another . E Draw EFHG cutting AB , KL , and CD , in F , H , and G. F A. B H Then EFB = FHL . Κ . L G And FHL = HGD . C D Therefore EFB ...
E. M. Reynolds. THEOREM IX . Straight lines AB , CD parallel to the same line KL are parallel to one another . E Draw EFHG cutting AB , KL , and CD , in F , H , and G. F A. B H Then EFB = FHL . Κ . L G And FHL = HGD . C D Therefore EFB ...
Σελίδα 19
... parallel . THEOREM XVI . The opposite sides and angles of parallelograms are equal . Let ABCD be a parallelogram . Then shall AB = CD , BC = DA , A B A = C , and B = D. Join BD . D Then the alternate angle ABD = the alternate angle CDB ...
... parallel . THEOREM XVI . The opposite sides and angles of parallelograms are equal . Let ABCD be a parallelogram . Then shall AB = CD , BC = DA , A B A = C , and B = D. Join BD . D Then the alternate angle ABD = the alternate angle CDB ...
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A'BC Algebra alternate angle altitude angle equal arc CD ARITHMETIC bisect BROOKE FOSS WESTCOTT Cambridge centre circle touching circumference Clifton College cloth coincide congruent CONIC SECTIONS contain convex angle convex polygon Crown 8vo diagonals diameter dicular divide EDUCATIONAL BOOKS Edward Thring ELEMENTARY TREATISE equal angles equally distant equally remote equiangular equilateral triangle Examples exterior angle Extra fcap Fcap Find the locus GEOMETRY given angle given circle given line given point given straight line Grammar greater Head Master Hence intersect isosceles John's College late Fellow Let ABCD line drawn measure meet middle point opposite sides parallel to CD parallelogram perpen perpendicular polygon produced quadrilateral figure radius rectangle right-angled triangle Schools Second Edition segments Shew shortest line similar triangles square on BC student tangent THEOREM THEOREM VII trapezium triangle ABC vertex