Now since, as C is to D, so is E to F, therefore, alternately, as C is to E, so is D to F. [VII. 13]
And, since A is plane, and C, D are its sides, therefore D by multiplying C has made A.
For the same reason also E by multiplying F has made B.
Now let D by multiplying E make G.
Then, since D by multiplying C has made A, and by multiplying E has made G,
therefore, as C is to E, so is A to G. [VII. 17]
But, as C is to E, so is D to F;
therefore also, as D is to F, so is A to G.
Again, since E by multiplying D has made G, and by multiplying F has made B,
therefore, as D is to F, so is G to B. [VII. 17]
But it was also proved that, as D is to F, so is A to G;
therefore also, as A is to G, so is G to B.
Therefore A, G, B are in continued proportion.
Therefore between A, B there is one mean proportional number.