If four numbers are in continued proportion, and the first is a cube, then the fourth is also a cube.

Let *A, B, C,* and *D* be four numbers in continued proportion, and let *A* be a cube.

I say that *D* is also a cube.

Since between *A* and *D* there are two mean proportional numbers *B* and *C,* therefore *A* and *D* are similar solid numbers. But *A* is a cube, therefore *D* is also a cube.

Therefore, *if four numbers are in continued proportion, and the first is a cube, then the fourth is also a cube.*

Q.E.D.