Let A the first be the same multiple of B the second, that C the third is of D the fourth; and of A and C let the equimultiples EF and GH be taken: EF shall be the same multiple of B that GH is of D.
For, because EF is the same multiple of A that GH is of D, [Hypothesis.
as many magnitudes as there are in EF equal to A, so many are there in GH equal to C.
Divide EF into the magnitudes EK, KF, each equal to A; and GH into the magnitudes GL, LH, each equal to C.
Therefore the number of the magnitudes EK,KL, will be equal to the number of the magnitudes GL, LH.
And because A is the same multiple of B that C of D, [Hypothesis
and that EK is equal to A and GL is equal to C; [Constr
therefore EK is the same multiple of B that GL is of D.
For the same reason KF is the same multiple of B that LH is of D.
Therefore because EK the first is the same multiple of B the second, that GL the third is of D the fourth,
and that KF the fifth is the same multiple of B the second that LH the sixth is of D the fourth;
EF the first together with the fifth, is the same multiple of B the second, that GH the third together with the sixth, is of D the fourth. [V. 2.
In the same manner, if there be more parts in EF equal to A and in GH equal to C, it may be shewn that EF is the same multiple of B that GH is of D. [V. 2, Cor.
Wherefore, if the first &c. q.e.d.
PROPOSITION 4. THEOREM.
If the first have the same ratio to the second that the third has to the fourth and if there be taken any equi-
