Euclidian GeometryMacmillan, 1874 - 349 sider |
Inni boken
Resultat 1-5 av 49
Side xvi
... , for divides read divide . ,, 129 , " , 1 , add if possible , let them bisect each other ; " IZI ,,, 13 , omit to it . " " 201. In the diagram the diagonal of FB is AC . 2 PLANE GEOMETRY . BOOK I. INTRODUCTION . A point xvi.
... , for divides read divide . ,, 129 , " , 1 , add if possible , let them bisect each other ; " IZI ,,, 13 , omit to it . " " 201. In the diagram the diagonal of FB is AC . 2 PLANE GEOMETRY . BOOK I. INTRODUCTION . A point xvi.
Side 11
... AB is = AC , ' .. AB is = BC , .. 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 . Let DAE STRAIGHT LINES , ANGLES AND TRIANGLES . II PROBLEM A. ...
... AB is = AC , ' .. AB is = BC , .. 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 . Let DAE STRAIGHT LINES , ANGLES AND TRIANGLES . II PROBLEM A. ...
Side 12
Francis Cuthbertson. PROBLEM B. To bisect a given angle . 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 circles ...
Francis Cuthbertson. PROBLEM B. To bisect a given angle . 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 circles ...
Side 13
Francis Cuthbertson. PROBLEM C. To bisect a given straight line . 72 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. and Join CD ...
Francis Cuthbertson. PROBLEM C. To bisect a given straight line . 72 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. and 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 ; - ( 1. 6 ) .. the EAB is the = ECF . ( I. 1 ) But the ACD is > the △ ECF ; LACD is the EAB ...
... 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 ; - ( 1. 6 ) .. the EAB is the = ECF . ( I. 1 ) But the ACD is > the △ ECF ; LACD is the EAB ...
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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 Examples 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 revised 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