Let, then, AK, KH, HB be divisions which are equal in multitude with DF, FG, GE.
Now, since DE is greater than AB, and from DE there has been subtracted EG less than its half, and, from AB, BH greater than its half, therefore the remainder GD is greater than the remainder HA.
And, since GD is greater than HA, and there has been subtracted, from GD, the half GF, and, from HA, HK greater than its half, therefore the remainder DF is greater than the remainder AK.
But DF is equal to C;
therefore C is also greater than AK.
Therefore AK is less than C.