The Elements of Geometry: Or, The First Six Books, with the Eleventh and Twelfth, of Euclid |
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Seite 61
In like manner , it may be shown that FD is greater than CD . Therefore , D A is the greatest ; DE is greater than D F , and DF greater than D C. Because M K and KD are greater than MD ( I. 20 ) and MK is equal ( I. Def . 15 ) to MG .
In like manner , it may be shown that FD is greater than CD . Therefore , D A is the greatest ; DE is greater than D F , and DF greater than D C. Because M K and KD are greater than MD ( I. 20 ) and MK is equal ( I. Def . 15 ) to MG .
Seite 114
But it was shown that LM is the same multiple of CD , that G K is of A E. Therefore G K is the same multiple of A E , that L N is of CF ; that is , G K and LN are equimultiples of A E and CF. But KO and NP are equimultiples of BE and DF ...
But it was shown that LM is the same multiple of CD , that G K is of A E. Therefore G K is the same multiple of A E , that L N is of CF ; that is , G K and LN are equimultiples of A E and CF. But KO and NP are equimultiples of BE and DF ...
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be the circumference K B. Therefore the straight line K B is less than G U. Because the angle BZK is obtuse , BK is greater than BZ ; in the triangles AGU and A B Z , it may be shown , as in Prop . XV . , Book III . , that A Z is ...
be the circumference K B. Therefore the straight line K B is less than G U. Because the angle BZK is obtuse , BK is greater than BZ ; in the triangles AGU and A B Z , it may be shown , as in Prop . XV . , Book III . , that A Z is ...
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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 touch triangle triangle ABC twice vertex Vols Wherefore whole