A history of Greek mathematics - Wilbourhall.org

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ON THE CUTTTNG-OFF OF A RATIO 177
Draw OG parallel to BN or B'N' to meet AM in G.
D on AM such that OC:AB =X = B'N' : J.lf,
Then AM : ^D = .S'iV' : 00
= B'M:CM;
therefore MB:AB = B'G :
or CM .MB = AB . BV,
CM,
a given rectangle.
Take
Hence the problem is reduced to one of applying to GB a
MB) equal to a given rectangle (AB . B'G) but
rectangle (GM .
falling short by a square figure. In the case as drawn, whatever
be the value of A, the solution is always possible because
the given rectangle AB . GB' is always less than CA . AB, and
therefore always less than J-
GB 2 ;
one of the positions of
M falls between A and B because GM . MB