(diagram 1) Let there be two unequal magnitudes AB, G where AB is larger.  I say that if more than half is taken from AB, and more than half from what remains, and this repeatedly occurs, some magnitude will be left which is less than magnitude G.

(diagram 2) For G being multiplied  at some point will be larger than AB.  Let it be multiplied and let DE be a multiple of G yet larger than AB, and let DE be divided into DZ, ZH, HE equal to G, and let more than half, namely BQ, be taken from AB, and let more than half, namely QK, be taken from AQ, and let this repeatedly occur, until the divisions in AB become equinumerous with the divisions in DE.

(diagram 3) And so let AK, KQ, QB be divisions  which are equinumerous with DZ, ZH, HE.  And since DE is larger than AB, and less than half, namely EH, is taken from DE, while more than half, namely BQ, from AB, therefore the remainder HD is larger than the remainder QA.  And since HD is larger than QA, and half, HZ, is taken from HD, while more than half, QK, from QA, therefore the remainder DZ is larger than the remainder AK.  But DZ is equal to G.  Therefore, G is larger than AK.  Therefore AK is less than G.

Therefore, magnitude AK is left of the AB magnitude, although it is less than the displayed smaller magnitude G, which it was required to prove.  It will be proved similarly even if what get taken away are half.