The Elements of Geometry: Or, The First Six Books, with the Eleventh and Twelfth, of Euclid |
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Seite 46
But the rect- angle N B.BD and the square of CB is equal ( II . 5 ) to the square of CD . And it has been proved that A D is equal to NB . Therefore the rectangle A D.D B , and the square of CB is equal to the square of CD . PROP .
But the rect- angle N B.BD and the square of CB is equal ( II . 5 ) to the square of CD . And it has been proved that A D is equal to NB . Therefore the rectangle A D.D B , and the square of CB is equal to the square of CD . PROP .
Seite 50
Because E C is equal to CA , the square of EC is equal to the square of CA. Therefore the squares of EC and CA are double of the square of CA. But the square of E A is equal ( I. 47 ) to the squares of EC and CA. Therefore the square of ...
Because E C is equal to CA , the square of EC is equal to the square of CA. Therefore the squares of EC and CA are double of the square of CA. But the square of E A is equal ( I. 47 ) to the squares of EC and CA. Therefore the square of ...
Seite 192
Because the square of BD is to the square of FH ( XII . 1 ) as the polygon AXBOCPDR is to the polygon EKFLGMHN . But the square of BD is to the square of FH ( Hyp . ) as the circle A B C D is to the space S. Therefore the circle A B C D ...
Because the square of BD is to the square of FH ( XII . 1 ) as the polygon AXBOCPDR is to the polygon EKFLGMHN . But the square of BD is to the square of FH ( Hyp . ) as the circle A B C D is to the space S. Therefore the circle A B C D ...
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The Elements of Geometry: Or, the First Six Books, with the Eleventh and ... Euclides Keine Leseprobe verfügbar - 2016 |
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A B C ABCD altitude applied base bisected Book Cassell's centre circle circumference cloth coincide common cone Const contained Corollary cylinder definition demonstration described diagonal diameter divided double draw Edition equal equal angles equiangular equimultiples Exercise extremities fore four fourth given straight line greater half Illustrated inscribed interior join less magnitudes manner meet multiple parallel parallelogram parallelopiped pass perpendicular plane polygon prism PROBLEM produced PROP proportionals proposition proved pyramid ratio reason rectangle rectangle contained rectilineal figure remaining angle right angles segment shown sides similar solid solid angle sphere square straight lines A B taken THEOREM third triangle triangle ABC twice vertex Vols Wherefore whole