To find the third apotome.

Set out a rational straight line *A*. Set out three numbers *E*, *BC*, and *CD* which do not have to one another the ratio which a square number has to a square number, but let *CB* have to *BD* the ratio which a square number has to a square number.

Let it be contrived that *E* is to *BC* as the square on *A* is to the square on *FG*, and *BC* is to *CD* as the square on *FG* is to the square on *GH*.

Since *E* is to *BC* as the square on *A* is to the square on *FG*, therefore the square on *A* is commensurable with the square on *FG*.

But the square on *A* is rational, therefore the square on *FG* is also rational, therefore *FG* is rational.

Since *E* does not have to *BC* the ratio which a square number has to a square number, therefore neither has the square on *A* to the square on *FG* the ratio which 3 square number has to a square number. Therefore *A* is incommensurable in length with *FG*.

Again, since *BC* is to *CD* as the square on *FG* is to the square on *GH*, therefore the square on *FG* is commensurable with the square on *GH*.

But the square on *FG* is rational, therefore the square on *GH* is also rational, therefore *GH* is rational.

Since *BC* does not have to *CD* the ratio which a square number has to a square number, therefore neither has the square on *FG* to the square on *GH* the ratio which a square number has to a square number. Therefore *FG* is incommensurable in length with *GH*.

And both are rational, therefore *FG* and *GH* are rational straight lines commensurable in square only. Therefore *FH* is an apotome.

I say next that it is also a third apotome.

Since *E* is to *BC* as the square on *A* is to the square on *FG*, and *BC* is to *CD* as the square on *FG* is to the square on *HG*, therefore, *ex aequali*, *E* is to *CD* as the square on *A* is to the square on *HG*.

But *E* does not have to *CD* the ratio which a square number has to a square number, therefore neither has the square on *A* to the square on *GH* the ratio which a square number has to a square number. Therefore *A* is incommensurable in length with *GH*.

Therefore neither of the straight lines *FG* nor *GH* is commensurable in length with the rational straight line *A* set out.

Now let the square on *K* be that by which the square on *FG* is greater than the square on *GH*.

Since *BC* is to *CD* as the square on *FG* is to the square on *GH*, therefore, in conversion, *BC* is to *BD* as the square on *FG* is to the square on *K*.

But *BC* has to *BD* the ratio which a square number has to a square number, therefore the square on *FG* also has to the square on *K* the ratio which a square number has to a square number.

Therefore *FG* is commensurable in length with *K*, and the square on *FG* is greater than the square on *GH* by the square on a straight line commensurable with *FG*.

And neither of the straight lines *FG* nor *GH* is commensurable in length with the rational straight line *A* set out, therefore *FH* is a third apotome.

Therefore the third apotome *FH* has been found.

Q.E.F.