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If a first magnitude have to a second the same ratio as a third has to a fourth, and the first be greater than the third, the second will also be greater than the fourth; if equal, equal; and if less, less.

For let a first magnitude A have the same ratio to a second B as a third C has to a fourth D; and let A be greater than C; I say that B is also greater than D.

For, since A is greater than C, and B is another, chance, magnitude, therefore A has to B a greater ratio than C has to B. [V. 8]

But, as A is to B, so is C to D;

therefore C has also to D a greater ratio than C has to B. [V. 13]

But that to which the same has a greater ratio is less; [V. 10]

therefore D is less than B; so that B is greater than D.

Similarly we can prove that, if A be equal to C, B will also be equal to D; and, if A be less than C, B will also be less than D.

Therefore etc. Q. E. D.

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