If three straight lines are proportional, then the rectangle contained by the extremes equals the square on the mean; and, if the rectangle contained by the extremes equals the square on the mean, then the three straight lines are proportional.

Let the three straight lines *A* and *B* and *C* be proportional, so that *A* is to *B* as *B* is to *C.*

I say that the rectangle *A* by *C* equals the square on *B.*

Make *D* equal to *B.*

Then, since *A* is to *B* as *B* is to *C,* and *B* equals *D,* therefore *A* is to *B* as *D* is to *C.*

But, if four straight lines are proportional, then the rectangle contained by the extremes equals the rectangle contained by the means.

Therefore the rectangle *A* by *C* equals the rectangle *B* by *D.* But the rectangle *B* by *D* is the square on *B,* for *B* equals *D,* therefore the rectangle *A* by *C* equals the square on *B.*

Next, let the rectangle *A* by *C* equal the square on *B.*

I say that *A* is to *B* as *B* is to *C.*

With the same construction, since the rectangle *A* by *C* equals the square on *B,* while the square on *B* is the rectangle *B* by *D,* for *B* equals *D,* therefore the rectangle *A* by *C* equals the rectangle *B* by *D.*

But, if the rectangle contained by the extremes equals that contained by the means, then the four straight lines are proportional.

Therefore *A* is to *B* as *D* is to *C.*

But *B* equals *D,* therefore *A* is to *B* as *B* is to *C.*

Therefore, *if three straight lines are proportional, then the rectangle contained by the extremes equals the square on the mean; and, if the rectangle contained by the extremes equals the square on the mean, then the three straight lines are proportional.*

Q.E.D.