Between two square numbers there is one mean proportional number, and the square has to the square the duplicate ratio of that which the side has to the side.

Let *A* and *B* be square numbers, and let *C* be the side of *A,* and *D* of *B.*

I say that between *A* and *B* there is one mean proportional number, and *A* has to *B* the ratio duplicate of that which *C* has to *D.*

Multiply *C* by *D* to make *E.*

Now, since *A* is a square and *C* is its side, therefore *C* multiplied by itself makes *A.* For the same reason also, *D* multiplied by itself makes *B.*

Since, then, *C* multiplied by the numbers *C* and *D* makes *A* and *E* respectively, therefore *C* is to *D* as *A* is to *E.*

For the same reason also *C* is to *D* as *E* is to *B.* Therefore *A* is to *E* as *E* is to *B.* Therefore between *A* and *B* there is one mean proportional number.

I say next that *A* also has to *B* the ratio duplicate of that which *C* has to *D.*

Since *A, E,* and *B* are three numbers in proportion, therefore *A* has to *B* the ratio duplicate of that which *A* has to *E.*

But *A* is to *E* as *C* is to *D,* therefore *A* has to *B* the ratio duplicate of that which the side *C* has to *D.*

Therefore, *between two square numbers there is one mean proportional number, and the square has to the square the duplicate ratio of that which the side has to the side.*

Q.E.D.