For let GH be made equal to AB.
Therefore, whatever parts GH is of CD, the same parts also is AE of CF.
Let GH be divided into the parts of CD, namely GK, KH, and AE into the parts of CF, namely AL, LE;
thus the multitude of GK, KH will be equal to the multitude of AL, LE.
Now since, whatever part GK is of CD, the same part also is AL of CF, while. CD is greater than CF,
therefore GK is also greater than AL.
Let GM be made equal to AL.
Therefore, whatever part GK is of CD, the same part also is GM of CF;
therefore also the remainder MK is the same part of the remainder FD that the whole GK is of the whole CD. [VII. 7]
Again, since, whatever part KH is of CD, the same part also is EL of CF, while CD is greater than CF,
therefore HK is also greater than EL.
Let KN be made equal to EL.
Therefore, whatever part KH is of CD, the same part also is KN of CF;
therefore also the remainder NH is the same part of the remainder FD that the whole KH is of the whole CD. [VII. 7]
But the remainder MK was also proved to be the same part of the remainder FD that the whole GK is of the whole CD;
therefore also the sum of MK, NH is the same parts of DF that the whole HG is of the whole CD.
But the sum of MK, NH is equal to EB, and HG is equal to BA;
therefore the remainder EB is the same parts of the remainder FD that the whole AB is of the whole CD.
Q. E. D.
Complete text
Book I