The first six books of the Elements of Euclid, with numerous exercises |
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Σελίδα 6
to c , and from a b , the greater of two straight lines , a part a e has been cut off equal to c the less . ... because the angle bac is equal to the angle e df ; wherefore also the point c shall coincide with the point f , because the ...
to c , and from a b , the greater of two straight lines , a part a e has been cut off equal to c the less . ... because the angle bac is equal to the angle e df ; wherefore also the point c shall coincide with the point f , because the ...
Σελίδα 7
09 In bd take any point f , and from a e the greater , cut off ag equal ( i . 3 ) to a f , the less , and join fc , g b . Because a f is equal to ag , and a b to a c , the two sides fa , a c are equal to the two ga , a b , each to each ...
09 In bd take any point f , and from a e the greater , cut off ag equal ( i . 3 ) to a f , the less , and join fc , g b . Because a f is equal to ag , and a b to a c , the two sides fa , a c are equal to the two ga , a b , each to each ...
Σελίδα 8
But if one of the vertices , as d , be within the other triangle a cb ; produce a c , ad to e , f ; therefore , because a c is equal to ad in the triangle acd , the angles ecd , fdc upon the other side of the base cd are equal ( i .
But if one of the vertices , as d , be within the other triangle a cb ; produce a c , ad to e , f ; therefore , because a c is equal to ad in the triangle acd , the angles ecd , fdc upon the other side of the base cd are equal ( i .
Σελίδα 9
the two sides de , d f , each to each , viz . ab to d e , and a c to df ; and le also the base b c equal to the base e f . The angle bac is a d 8 equal to the angle ed f . For , if the triangle a b c be applied to def , so that the ...
the two sides de , d f , each to each , viz . ab to d e , and a c to df ; and le also the base b c equal to the base e f . The angle bac is a d 8 equal to the angle ed f . For , if the triangle a b c be applied to def , so that the ...
Σελίδα 11
the circle egf meeting a b in f , g ; and bisect ( i . 10 ) fg in h , and join h cf , ch , cg ; the straight a f g b line ch , drawn from the d given point c , is perpendicular to the given straight line a b . Because fh is equal to hg ...
the circle egf meeting a b in f , g ; and bisect ( i . 10 ) fg in h , and join h cf , ch , cg ; the straight a f g b line ch , drawn from the d given point c , is perpendicular to the given straight line a b . Because fh is equal to hg ...
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a b c angle a b c angle bac base base bc bc is equal bisected centre circle circumference common compounded definition demonstrated describe diameter divided double draw equal angles equal to f equiangular equilateral equimultiples exterior angle extremity fall fore four fourth given straight line greater half ILLUSTRATED inscribe join less likewise magnitudes manner meet multiple parallel parallelogram pass perpendicular produced proportionals PROPOSITION ratio reason rectangle contained rectilineal figure remaining angle right angles segment shewn sides similar square square of a c straight line a b Take taken third touches the circle triangle triangle a b c wherefore whole