Square on Rational Straight Line applied to Binomial Straight Line

Proof

 * Euclid-X-112.png

Let $A$ be a rational straight line.

Let $BC$ be a binomial.

Let $DC$ be the greater term of $BC$.

Let $BC \cdot EF$ be the rectangle on $BC$ equal to $A^2$.

It is to be demonstrated that:
 * $EF$ is an apotome whose terms are commensurable with $CD$ and $DB$ and in the same ratio

and:
 * the order of $EF$ is the same as the order of $BC$.

Let $BD \cdot G$ be the rectangle on $BD$ equal to $A^2$.

Then:
 * $BC \cdot EF = BD \cdot G$

So from :
 * $CB : BD = G : EF$

But:
 * $CB > BD$

and so from:

and:

it follows that:
 * $G > EF$

Let $EH = G$.

Then:
 * $CB : BD = HE : EF$

So from :
 * $CD : BD = HF : FE$

Let it be contrived that $HF : FE = FK : KE$.

Therefore from :
 * $HK : KF = FK : KE$

Therefore from :
 * $FK : KE = CD : DB$

and:
 * $HK : KF = CD : DB$

But from the definition of binomial, $CD^2$ is commensurable with $DB^2$.

Therefore from:

and:

it follows that:
 * $HK^2$ is commensurable with $KF^2$.

We have that $HK$, $KF$ and $KE$ are proportional.

So:
 * $HK^2 : KF^2 = HK : KE$

Thus $HK$ is commensurable in length with $KE$.

Therefore from :
 * $HE$ is commensurable in length with $EK$.

We have that $A^2 = EH \cdot BD$.

But $A^2$ is rational.

Therefore $EH \cdot BD$ is also rational.

We have that $EH \cdot BD$ is applied to the rational straight line $BD$.

Therefore from :
 * $EH$ is rational and commensurable in length with $BD$.

We have that $EK$ is rational and commensurable in length with $EH$.

So $EK$ is rational and commensurable in length with $BD$.

We have that:
 * $CD : DB = FK : KE$

and:
 * $CD$ and $DB$ are straight lines which are commensurable in square only.

Therefore by :
 * $FK$ and $KE$ are straight lines which are commensurable in square only.

But $KE$ is rational.

Therefore $FK$ is also rational.

Therefore $FK$ and $KE$ are rational straight lines which are commensurable in square only.

Therefore $EF$ is an apotome.

We have that:
 * $CD^2 = DB^2 + \lambda^2$

where either:
 * $\lambda$ is commensurable in length with $CD$

or:
 * $\lambda$ is incommensurable in length with $CD$.

First suppose $\lambda$ is commensurable in length with $CD$.

Then by :
 * $FK^2 = KE^2 + \mu^2$

where $\mu$ is commensurable in length with $FK$.

Let $CD$ be commensurable in length with a rational straight line $\alpha$ which has been set out.

Then by:

and:

it follows that:
 * $FK$ is commensurable in length with $\alpha$.

Let $BD$ be commensurable in length with $\alpha$.

Then by :
 * $KE$ is commensurable in length with $\alpha$.

Let neither $CD$ nor $BD$ be commensurable in length with $\alpha$.

Then neither $FK$ nor $KE$ is commensurable in length with $\alpha$.

Next suppose $\lambda$ is incommensurable in length with $HF$.

Then by :
 * $FK^2 = KE^2 + \mu^2$

where $\mu$ is incommensurable in length with $FK$.

Let $CD$ be commensurable in length with a rational straight line $\alpha$ which has been set out.

Then by:

and:

it follows that:
 * $FK$ is commensurable in length with $\alpha$.

Let $BD$ be commensurable in length with $\alpha$.

Then by :
 * $KE$ is commensurable in length with $\alpha$.

Let neither $CD$ nor $BD$ be commensurable in length with $\alpha$.

Then neither $FK$ nor $KE$ is commensurable in length with $\alpha$.

It follows that:
 * the terms of $EF$ are commensurable with the terms of $BD$ and in the same ratio

and:
 * the order of $EF$ is the same as the order of $BD$.