Thirdly, let A be less than C: B shall be less than D.
For, C is greater than A.
And because C is to D as A is to B; [Hypothesis.
and C is greater than A;
therefore, by the first case, D is greater than B;
that is, B is less than D.
Wherefore, if the first &c. q.e.d.
PROPOSITION 15. THEOREM.
Magnitudes have the same ratio to one another that their equimultiples have.
Let AB be the same multiple of C that DE is of F: C shall be to F as AB is to DE.
For, because AB is the same multiple of C that DE is of F, [Hypothesis.
therefore as many magnitudes as there are in AB equal to C, so many are there in DE equal to F.
Divide AB into the magnitudes AG, GH, HB, each equal to F; and DE into the magnitudes DK, KL, LE, each equal to F. Therefore the number of the magnitudes AG, GH, HB will be equal to the number of the magnitudes DK, KL, LE.
And because AG', GH, HB are all equal; [Construction.
and that DK, KL, LE are also all equal;
therefore AG is to DK as GH is to KL, and as HB is to LE. [V. 7.
But as one of the antecedents is to its consequent, so are all the antecedents to all the consequents. [V. 12.
Therefore as AG is to DK so is AB to DE.
But AG equal to C, and DK is equal to F.
Therefore as C is to F so is AB to DE.
Wherefore, magnitudes &c. q.e.d.
