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Book III   Proposition 24
For, if the straight line AB coincide with CD but the segment AEB does not coincede with CFD,
    it will either fall within it [as CGD], or outside it: which is impossible [III.23]
or
    it will fall awry, as CGD, and a circle cuts a circle at more points than two: which is impossible. [III.10]

therefore, if the straight line AB be applied to CD,
    the segment AEB will not fail to coincide with CFD also;
therefore it will coincide with it and will be equal to it.

Being what it was required to prove.
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