Then, since CF is equal to FE, [I. 43] and FB is common,
therefore the whole CH is equal to the whole KE.
But CH is equal to CG, since AC is also equal to CB. [I. 36]
Therefore GC is also equal to EK.
Let CF be added to each;
therefore the whole AF is equal to the gnomon LMN;
so that the parallelogram DB, that is, AD, is greater than the parallelogram AF.