Euclidian Geometry |
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Resultat 1-5 av 62
Side 10
... taken for granted that a circle can be described having any centre and radius equal to any given straight line . PROBLEM A. Describe an equilateral triangle upon a given straight IO STRAIGHT LINES , ANGLES AND TRIANGLES .
... taken for granted that a circle can be described having any centre and radius equal to any given straight line . PROBLEM A. Describe an equilateral triangle upon a given straight IO STRAIGHT LINES , ANGLES AND TRIANGLES .
Side 11
... describe an equilateral triangle upon AB . With A as centre , and radius AB , describe the circle BCD , and with B as centre and radius BA describe the circle ACE . and Let these circles intersect in C ; join CA , CB . Then ABC shall be ...
... describe an equilateral triangle upon AB . With A as centre , and radius AB , describe the circle BCD , and with B as centre and radius BA describe the circle ACE . and Let these circles intersect in C ; join CA , CB . Then ABC shall be ...
Side 12
... describe a circle cutting AD , AE in the points B and C. With centres B and C describe equal circles cutting one another in F Join AF ; then △ DAE shall be bisected by AF . Then = Join BF , CF. in the As ABF , ACF , BA is CA , BF = CF ...
... describe a circle cutting AD , AE in the points B and C. With centres B and C describe equal circles cutting one another in F Join AF ; then △ DAE shall be bisected by AF . Then = Join BF , CF. in the As ABF , ACF , BA is CA , BF = CF ...
Side 13
... describe two circles having equal radii 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 ...
... describe two circles having equal radii 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 ...
Side 16
... describe equal Os 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 ) ...
... describe equal Os 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 ) ...
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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