The Elements of Euclid |
Dentro del libro
Resultados 1-3 de 91
Página 46
... becaufe BA is equal to AD , being fides of a square ; wherefore the angle CGB is equal to the angle GBC , and therefore the fide BC is equal to the fide CG . but CB is equal alfo f to GK , and CG to BK ; wherefore the figure CGKB is ...
... becaufe BA is equal to AD , being fides of a square ; wherefore the angle CGB is equal to the angle GBC , and therefore the fide BC is equal to the fide CG . but CB is equal alfo f to GK , and CG to BK ; wherefore the figure CGKB is ...
Página 155
... as therefore ED to DF , fo is B GD to DF ; wherefore ED is c b . ... are equal to the two fides GD , DF ; and the angle EDF is equal to the angle GDF , wherefore . the bafe EF is equal to the bafe FG f , and the triangle EDF to f .
... as therefore ED to DF , fo is B GD to DF ; wherefore ED is c b . ... are equal to the two fides GD , DF ; and the angle EDF is equal to the angle GDF , wherefore . the bafe EF is equal to the bafe FG f , and the triangle EDF to f .
Página 370
Dat . given a . make the ratio of AD to DE the fame with this ratio ; therefore the ratio of AD to DE is given . and AD is given , wherefore A b . 4. Dat . b DE , and the remainder AE are gi- e . 12. 5 . EDB с C ven . and because as DC ...
Dat . given a . make the ratio of AD to DE the fame with this ratio ; therefore the ratio of AD to DE is given . and AD is given , wherefore A b . 4. Dat . b DE , and the remainder AE are gi- e . 12. 5 . EDB с C ven . and because as DC ...
Comentarios de la gente - Escribir un comentario
No encontramos ningún comentario en los lugares habituales.
Otras ediciones - Ver todas
Términos y frases comunes
added alfo alſo altitude angle ABC angle BAC bafe baſe becauſe bifected Book Book XI circle circle ABCD circumference common cone cylinder defcribed Definition demonftrated diameter divided double draw drawn equal equal angles equiangular equimultiples excefs fame fame multiple fecond fegment fhall fhewn fides fimilar firft firſt folid folid angle fore four fourth fquare fquare of BC ftraight line given angle given ftraight line given in fpecies given in pofition given magnitude given ratio greater Greek half join lefs magnitude manner meet oppofite parallel parallelogram perpendicular plane produced PROP Propofition proportionals pyramid reafon rectangle remaining right angles ſphere taken THEOR theſe third thro triangle ABC wherefore whole