Euclidian Geometry |
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Resultat 1-5 av 49
Side xvi
... , 129 ,,, 6 , for divides read divide . I , add if possible , let them bisect each other ; ,, IZI ,,, 13 , omit to it . " " 201. In the diagram the diagonal of FB is AC . PLANE GEOMETRY . BOOK I. INTRODUCTION . A point has xvi.
... , 129 ,,, 6 , for divides read divide . I , add if possible , let them bisect each other ; ,, IZI ,,, 13 , omit to it . " " 201. In the diagram the diagonal of FB is AC . PLANE GEOMETRY . BOOK I. INTRODUCTION . A point has xvi.
Side 11
... centre of the · B is the centre of the ACE , .. AC is BC ; .. ABC is an equilateral triangle , and it has been described upon AB . Q. E. F. PROBLEM B. To bisect a given angle . E Let STRAIGHT LINËS , ANGLES AND TRIANGLES . I I.
... centre of the · B is the centre of the ACE , .. AC is BC ; .. ABC is an equilateral triangle , and it has been described upon AB . Q. E. F. PROBLEM B. To bisect a given angle . E Let STRAIGHT LINËS , ANGLES AND TRIANGLES . I I.
Side 12
Francis Cuthbertson (M.A.). PROBLEM B. To bisect a given angle . E Let DAE be the given angle . It is required to bisect it . With centre A describe a circle cutting AD , AE in the points B and C. With centres B and C describe equal ...
Francis Cuthbertson (M.A.). PROBLEM B. To bisect a given angle . E Let DAE be the given angle . It is required to bisect it . With centre A describe a circle cutting AD , AE in the points B and C. With centres B and C describe equal ...
Side 13
Francis Cuthbertson (M.A.). PROBLEM C. To bisect a given straight line . Let AB be the given straight line . It is required to bisect it . With centres A and B describe two circles having equal radii intersecting in C , D. Join CD ...
Francis Cuthbertson (M.A.). PROBLEM C. To bisect a given straight line . Let AB be the given straight line . It is required to bisect it . With centres A and B describe two circles having equal radii intersecting in C , D. Join CD ...
Side 22
... Bisect AC in E ; join BE and produce BE to F , making EF- BE . Join CF. Then , · . AE , EB are respectively = CE , EF , and △ AEB is = △ CEF ; .. the = EAB is the ECÈ ( 1. 6 ) ( I. I ) But the ACD is > the △ ECF ; .. △ ACD is > the ...
... Bisect AC in E ; join BE and produce BE to F , making EF- BE . Join CF. Then , · . AE , EB are respectively = CE , EF , and △ AEB is = △ CEF ; .. the = EAB is the ECÈ ( 1. 6 ) ( I. I ) But the ACD is > the △ ECF ; .. △ ACD is > the ...
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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