If a square measures a square, then the side also measures the side; and, if the side measures the side, then the square also measures the square.

Let *A* and *B* be square numbers, let *C* and *D* be their sides, and let *A* measure *B.*

I say that *C* also measures *D.*

as in VIII.11

Multiply *C* by *D* to make *E.* Then *A, E,* and *B* are continuously proportional in the ratio of *C* to *D.*

And, since *A, E,* and *B* are continuously proportional, and *A* measures *B,* therefore *A* also measures *E.* And *A* is to *E* as *C* is to *D,* therefore *C* measures *D.*

Next, let *C* measure *D.*

I say that *A* also measures *B.*

With the same construction, we can in a similar manner prove that *A, E,* and *B* are continuously proportional in the ratio of *C* to *D.* And since *C* is to *D* as *A* is to *E,* and *C* measures *D,* therefore *A* also measures *E.*

And *A, E,* and *B* are continuously proportional, therefore *A* also measures *B.*

Therefore, *if a square measures a square, then the side also measures the side; and, if the side measures the side, then the square also measures the square.*

Q.E.D.

This proposition is used to prove its contrapositive, VIII.16.