Area contained by Apotome and Binomial Straight Line Commensurable with Terms of Apotome and in same Ratio

Proof

 * Euclid-X-114.png

Let the rectangle $AB \cdot CD$ be contained by the apotome $AB$ and the binomial straight line $CD$.

Let $CE$ be the greater term of $CD$.

Let:
 * $CE$ be commensurable in length with $AF$
 * $ED$ be commensurable in length with $FB$
 * $CE : ED = AF : FB$

Let the "side" of the rectangle $AB \cdot CD$ be $G$.

It is to be demonstrated that $G$ is rational.

Let $H$ be a rational straight line.

Let a rectangle equal to $H^2$ be applied to $CD$ which produces $KL$ as breadth.

By definition, $KL$ is an apotome.

Let the terms of $KL$ be $KM$ and $ML$.

Let $KM$ and $ML$ be commensurable with the terms $CE$ and $ED$ of the binomial straight line $CD$.

But from :
 * $CE$ and $ED$ are commensurable with the terms $AF$ and $FB$ of the apotome $CD$.

Therefore:
 * $AF : FB = KM : ML$

and so:
 * $AF : KM = BF : LM$

and so by :
 * $AB : KL = AF : KM$

But from :
 * $AF$ is commensurable in length with $KM$.

Therefore from :
 * $AB$ is commensurable in length with $KL$.

From :
 * $AB : KL = CD \cdot AB : CD \cdot KL$

Therefore from :
 * $CD \cdot AB$ is commensurable with $CD \cdot KL$.

But $CD \cdot KL = H^2$.

Therefore $CD \cdot AB$ is commensurable with $H^2$.

But $G^2 = CD \cdot AB$.

Therefore $G^2$ is commensurable with $H^2$.

But $H^2$ is rational.

Therefore $G^2$ is also rational.

Therefore $G$ is rational straight line.

But $G^2$ is the "side" of the rectangle $AB \cdot CD$.

Hence the result.