#### PROPOSITION 114.

If an area be contained by an apotome and the binomial straight line the terms of which are commensurable with the terms of the apotome and in the same ratio, the “side” of the area is rational.

For let an area, the rectangle AB, CD, be contained by the apotome AB and the binomial straight line CD, and let CE be the greater term of the latter; let the terms CE, ED of the binomial straight line be commensurable with the terms AF, FB of the apotome and in the same ratio; and let the “side” of the rectangle AB, CD be G; I say that G is rational.

For let a rational straight line H be set out, and to CD let there be applied a rectangle equal to the square on H and producing KL as breadth.

Therefore KL is an apotome.

Let its terms be KM, ML commensurable with the terms CE, ED of the binomial straight line and in the same ratio. [X. 112]

But CE, ED are also commensurable with AF, FB and in the same ratio; therefore, as AF is to FB, so is KM to ML.

Therefore, alternately, as AF is to KM, so is BF to LM; therefore also the remainder AB is to the remainder KL as AF is to KM. [V. 19]

But AF is commensurable with KM; [X. 12] therefore AB is also commensurable with KL. [X. 11]

And, as AB is to KL, so is the rectangle CD, AB to the rectangle CD, KL; [VI. 1] therefore the rectangle CD, AB is also commensurable with the rectangle CD, KL. [X. 11]

But the rectangle CD, KL is equal to the square on H; therefore the rectangle CD, AB is commensurable with the square on H.

But the square on G is equal to the rectangle CD, AB; therefore the square on G is commensurable with the square on H.

But the square on H is rational; therefore the square on G is also rational; therefore G is rational.

And it is the “side” of the rectangle CD, AB.

Therefore etc.

#### PORISM.

And it is made manifest to us by this also that it is possible for a rational area to be contained by irrational straight lines. Q. E. D.