Now, since similar plane numbers are those which have their sides proportional, [[VII. Def. 21](elem.7.def.21)] therefore, as C is to D, so is E to F.

I say then that between A, B there is one mean proportional number, and A has to B the ratio duplicate of that which C has to E, or D to F, that is, of that which the corresponding side has to the corresponding side.

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](elem.7.13)]

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](elem.7.17)]

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](elem.7.17)]

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.

I say next that A also has to B the ratio duplicate of that which the corresponding side has to the corresponding side, that is, of that which C has to E or D to F.

For, since A, G, B are in continued proportion, A has to B the ratio duplicate of that which it has to G. [[V. Def. 9](elem.5.def.9)]

And, as A is to G, so is C to E, and so is D to F.

Therefore A also has to B the ratio duplicate of that which C has to E or D to F. Q. E. D.