Euclidian Geometry |
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Resultat 1-5 av 20
Side 11
... intersect in C ; join CA , CB . Then ABC shall be the equilateral triangle required . = BCD , .. AB is AC , .. AB is BC , ... A is the centre of the · B is the centre of the ACE , .. AC is BC ; .. ABC is an equilateral triangle , and it ...
... intersect in C ; join CA , CB . Then ABC shall be the equilateral triangle required . = BCD , .. AB is AC , .. AB is BC , ... A is the centre of the · B is the centre of the ACE , .. AC is BC ; .. ABC is an equilateral triangle , and it ...
Side 13
... intersecting in C , D. Join CD cutting AB in E ; then AB shall be bisected in E. Join AC , CB , BD , DA . Then in the As ACD , BCD , and AC , CD , DA are respectively = BC , CD , DB , ..AS ACD , BCD are equal in all respects , ..LACD is ...
... intersecting in C , D. Join CD cutting AB in E ; then AB shall be bisected in E. Join AC , CB , BD , DA . Then in the As ACD , BCD , and AC , CD , DA are respectively = BC , CD , DB , ..AS ACD , BCD are equal in all respects , ..LACD is ...
Side 14
... intersecting straight lines form the four angles A , B , H , K. Then shall the opposite angles A , B be one another . = For if the figure were taken up , reversed , and placed so that each of the arms of H might fall along the former ...
... intersecting straight lines form the four angles A , B , H , K. Then shall the opposite angles A , B be one another . = For if the figure were taken up , reversed , and placed so that each of the arms of H might fall along the former ...
Side 16
... intersect- ing in F. Join AF . Then AF shall be to BC . Join DF , EF . · DA is = EA , DF = EF , and AF common to the two AS DAF , EAF .. these as are equal in all respects ; ..L DAF is = LEAF , .. AF is to BC . ( I. 5 ) ( Def ...
... intersect- ing in F. Join AF . Then AF shall be to BC . Join DF , EF . · DA is = EA , DF = EF , and AF common to the two AS DAF , EAF .. these as are equal in all respects ; ..L DAF is = LEAF , .. AF is to BC . ( I. 5 ) ( Def ...
Side 20
... intersecting in F. Join AF , cutting BC in G. Then shall AG be 1 to BC . Join AD , AE , FD , FE . Then · . · • DA , AF , FD are respectively = EA , AF , FE ; .. AS DAF , EAF are equal in all respects , and ... △ DAF is = LEAF ( 1.5 ) ...
... intersecting in F. Join AF , cutting BC in G. Then shall AG be 1 to BC . Join AD , AE , FD , FE . Then · . · • DA , AF , FD are respectively = EA , AF , FE ; .. AS DAF , EAF are equal in all respects , and ... △ DAF is = LEAF ( 1.5 ) ...
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Algebra base Cambridge centre chord circumference cloth Conic Sections Crown 8vo Describe a circle diagonals diameter divided draw a straight ELEMENTARY TREATISE English equiangular equilateral Euclid Extra fcap fcap GEOMETRY given angle given circle given point given straight line Grammar greater H Let Hence inscribed intersecting isosceles triangle Latin Let ABC line bisecting locus Mathematical meet opposite angles Owens College parallel parallelogram perimeter perpendicular plane polygon PROBLEM produced Professor proportional PROPOSITION ratio rect rectangle rectangle contained rectilineal figure regular polygon respectively rhombus right angles Schools Second Edition segment similar Similarly squares on AC straight line drawn straight line joining tangent THEOREM TRIGONOMETRY twice rectangle twice the squares vertex