PROPOSITION 90.
To find the sixth apotome.
Let a rational straight line
A be set out, and three numbers
E,
BC,
CD not having to one another the ratio which a square number has to a square number; and further let
CB also not have to
BD the ratio which a square number has to a square number.
Let it be contrived that, as
E is to
BC, so is the square on
A to the square on
FG, and, as
BC is to
CD, so is the square on
FG to the square on
GH. [
X. 6, Por.]
Now since, as
E is to
BC, so is the square on
A to the square on
FG, therefore the square on
A is commensurable with the square on
FG. [
X. 6]
But the square on
A is rational; therefore the square on
FG is also rational; therefore
FG is also rational.
And, since
E has not 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 a square number has to a square number; therefore
A is incommensurable in length with
FG. [
X. 9]
Again, since, as
BC is to
CD, so is the square on
FG to the square on
GH, therefore the square on
FG is commensurable with the square on
GH. [
X. 6]
But the square on
FG is rational; therefore the square on
GH is also rational; therefore
GH is also rational.
And, since
BC has not to
CD the ratio which a square number has to a square number,
