The Elements of Geometry: Or, The First Six Books, with the Eleventh and Twelfth, of Euclid |
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Seite 103
For the same reason , M is the same multiple of B , that N is of D. Because , as A is to B , so is C to D ( Hyp . ) , and of A and C certain cquimultiples K and L have been taken , and of B and D certain equi- multiples M and N have ...
For the same reason , M is the same multiple of B , that N is of D. Because , as A is to B , so is C to D ( Hyp . ) , and of A and C certain cquimultiples K and L have been taken , and of B and D certain equi- multiples M and N have ...
Seite 129
For the same reason , D F is equal to FG . Because , in the triangles DEF and GEF , DE is equal to E G , and EF is common , the two sides D E and EF are equal to the two GE and E F , each to each . But the base D F is equal to the base ...
For the same reason , D F is equal to FG . Because , in the triangles DEF and GEF , DE is equal to E G , and EF is common , the two sides D E and EF are equal to the two GE and E F , each to each . But the base D F is equal to the base ...
Seite 140
For the same reason , the triangle E C D is similar to the triangle L H K. Therefore the similar polygons ABCDE and F G H KL are divided into the same number of similar triangles . Again , because the triangles are similar ...
For the same reason , the triangle E C D is similar to the triangle L H K. Therefore the similar polygons ABCDE and F G H KL are divided into the same number of similar triangles . Again , because the triangles are similar ...
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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