Elements of Geometry: Containing the First Six Books of Euclid, with a Supplement on the Quadrature of the Circle, and the Geometry of Solids; to which are Added, Elements of Plane and Spherical Trigonometry |
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Página 33
Because AD is equal to AE , and AF is common to the two triangles DAF , EAF ; the two sides DA , AF , are equal to the two D sides EA , AF , each to each ; but the base DF is also equal to the base EF ; thereHO fore the angle DAF is ...
Because AD is equal to AE , and AF is common to the two triangles DAF , EAF ; the two sides DA , AF , are equal to the two D sides EA , AF , each to each ; but the base DF is also equal to the base EF ; thereHO fore the angle DAF is ...
Página 42
therefore each of the opposite А B angles ABE , BED is a right angle ; whereВ fore the figure ADEB is rectangular , and it has been demonstrated that it is equilateral ; it is therefore a square , and it is described upon the given ...
therefore each of the opposite А B angles ABE , BED is a right angle ; whereВ fore the figure ADEB is rectangular , and it has been demonstrated that it is equilateral ; it is therefore a square , and it is described upon the given ...
Página 46
13. 1. ) AF parallel to CD or BE ; then AE = AD + CE . But AE = AB.BE = AB.BC , because BE = BC . So also AD = AC . be CD = AC.CB ; and CE = BC2 ; thereF D E fore AB.BC = AC.CB + BC2 . ca SCHOLIUM . In this proposition let AB be denoted ...
13. 1. ) AF parallel to CD or BE ; then AE = AD + CE . But AE = AB.BE = AB.BC , because BE = BC . So also AD = AC . be CD = AC.CB ; and CE = BC2 ; thereF D E fore AB.BC = AC.CB + BC2 . ca SCHOLIUM . In this proposition let AB be denoted ...
Página 77
to GC ; whereE fore AE.EC + ( 5.2 . ) ÈG = AG ?, and add- A C ing GF2 to both , AE.EC + EG + GF2 = AĞ ? + GF2 . Now EG2 + GF2 = EF2 , and B AG ? + GF = AF2 ; therefore AE.EC + EF2 = AF2 = FB2 But FB2 = BE.ED + ( 5 . 2. ) ...
to GC ; whereE fore AE.EC + ( 5.2 . ) ÈG = AG ?, and add- A C ing GF2 to both , AE.EC + EG + GF2 = AĞ ? + GF2 . Now EG2 + GF2 = EF2 , and B AG ? + GF = AF2 ; therefore AE.EC + EF2 = AF2 = FB2 But FB2 = BE.ED + ( 5 . 2. ) ...
Página 83
to AB , AC ; DF , EF produced will meet one another ; for , if they do not meet , they are parallel , whereА A А D E D D E B F C F o B В F B o fore , AB , AC , which are at right angles to them , are parallel , which is absurd ...
to AB , AC ; DF , EF produced will meet one another ; for , if they do not meet , they are parallel , whereА A А D E D D E B F C F o B В F B o fore , AB , AC , which are at right angles to them , are parallel , which is absurd ...
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Términos y frases comunes
ABCD altitude angle ABC angle BAC base bisected Book called centre chord circle circumference coincide common consequently construction cosine cylinder definition demonstrated described diameter difference distance divided double draw drawn equal equal angles equiangular equilateral Euclid exterior angle extremities fall fore four fourth given given straight line greater half Hence inscribed interior join less Let ABC magnitudes manner meet multiple opposite parallel parallelogram pass perpendicular plane polygon prism Prob produced PROP proportional proposition proved radius ratio reason rectangle contained rectilineal figure remaining right angles segment shewn sides similar sine solid square straight line taken tangent THEOR third touch triangle ABC wherefore whole