 (diagram 1) For let two magnitudes, AB, GD, be equal-times multiples of two magnitudes, E, Z, and [magnitudes] taken away, AH, GQ, be equal-times multiples of the same, E, Z. I say that remainders, HB, QD, are also either equal to E, Z, or are equal-times multiples of them. For let first HB be equal to E. I say that QD is also equal to Z. (diagram 2) For let GK be posited equal to Z. (diagram 3) Since AH is equally-times a multiple of E as GQ of Z, but HB is equal to E and KG to Z, therefore, AB is equally-times a multiple of E as KQ of Z. (v 2) But AB is supposed equally-times a multiple of E as GD of Z. Therefore, KQ is equally-times a multiple of Z as GD of Z. And so, since each of KQ, GD is equally-times a multiple of Z, therefore, KQ is equal to GD. (diagram 4) Let a common be taken away, GQ. Therefore, a remainder, KG, is equal to a remainder, QD. But Z is equal to KG. Therefore, QD is also equal to Z. Thus if HB is equal to E, QD will also be equal to Z. (diagram 5) Similarly, in fact, we will also show that if HB is a multiple of E, QD will also be so-much-a-multiple of Z. Therefore, if two magnitudes are equal-times multiples of two magnitudes, and some magnitudes taken away from them are equal-times multiples of the same, the remainders will be either equal to them or equal-times multiples of them, just what it was required to show.