Side of Regular Pentagon inscribed in Circle with Rational Diameter is Minor

Proof

 * Euclid-XIII-11.png

Let $ABCDE$ be a circle whose diameter is rational.

Let the regular pentagon $ABCDE$ be inscribed within the circle $ABCDE$.

It is to be demonstrated that the side of the pentagon $ABCDE$ is the irrational straight line called minor.

Let the center of the circle $ABCDE$ be $F$.

Let $AF$ be joined and produced to $G$.

Let $FB$ be joined and produced to $H$.

Let $AC$ be joined.

Let $K$ be placed so that $FK = \dfrac {AF} 4$.

We have that $AF$ is rational.

Therefore $FK$ is rational.

But $BF$ is rational.

Therefore $BK$ is rational.

We have that the arc $ACG$ equals the arc $ADG$.

Within those arcs, the arc $ABC$ equals the arc $AED$.

Therefore the remainders are equal: the arc $CG$ equals the arc $GD$.

If $AD$ is joined, it can be concluded that the angles at $L$ are right.

Also:
 * $CD = 2 \cdot CL$

For the same reason, the angles at $M$ are right.

Also:
 * $AC = 2 \cdot CM$

We have that:
 * $\angle ALC = \angle AMF$

and:
 * $\angle LAC$ is common to $\triangle ACL$ and $\triangle AMF$.

Therefore from :
 * $\angle ACL = \angle MFA$

Therefore:
 * $\triangle ACL$ is equiangular with $\triangle AMF$.

Therefore:
 * $LC : CA = MF : FA$

and so:
 * $2 \cdot LC : CA = 2 \cdot MF : FA$

But:
 * $2 \cdot MF : FA = MF : \dfrac {FA} 2$

Therefore:
 * $2 \cdot LC : CA = MF : \dfrac {FA} 2$

and so:
 * $2 \cdot LC : \dfrac {CA} 2 = MF : \dfrac {FA} 4$

But:
 * $DC = 2 \cdot LC$
 * $CM = \dfrac {CA} 2$
 * $FK = \dfrac {FA} 4$

Therefore:
 * $DC : CM = MF : FK$

Thus by :
 * $DC + CM : CM = MK : KF$

Therefore also:
 * $\left({DC + CM}\right)^2 : CM^2 = MK^2 : FK^2$

From :
 * if $AC$ is cut in extreme and mean ratio, the greater segment equals $CD$.

We have that:
 * $CM = \dfrac {AC} 2$

and so from :
 * $\left({DC + CM}\right)^2 = 5 \cdot CM^2$

But it was proved that:
 * $\left({DC + CM}\right)^2 : CM^2 = MK^2 : FK^2$

Therefore:
 * $MK^2 = 5 \cdot FK^2$

But $FK^2$ is rational.

Therefore $MK^2$ is rational.

We have that:
 * $BF = 4 \cdot FK$

Therefore:
 * $BK = 5 \cdot FK$

Therefore:
 * $BK^2 = 25 \cdot FK^2$

But:
 * $MK^2 = 5 \cdot FK^2$

Therefore:
 * $BK^2 = 5 \cdot MK^2$

Therefore the square on $BK$ has not to the square on $KM$ the ratio that a square number has to a square number.

Therefore from :
 * $BK$ is incommensurable in length with $KM$.

But each of $BK$ and $KM$ are rational.

Therefore $BK$ and $KM$ are rational straight lines which are commensurable in square only.

So $MB$ is a rational straight line from which another rational straight line with which it is commensurable in square only has been subtracted.

Therefore by definition:
 * $MB$ is an apotome

and:
 * $MK$ is the annex to $MB$.

It is now to be demonstrated that $MB$ is a fourth apotome.

Let $N$ be a magnitude such that:
 * $N^2 = BK^2 - KM^2$

Therefore:
 * $BK^2 = KM^2 + N^2$

We have that $KF$ is commensurable with $FB$.

From :
 * $KB$ is commensurable with $FB$.

But we have that $BF$ is commensurable with $BH$.

Therefore from :
 * $BK$ is commensurable with $BH$.

We have that:
 * $BK^2 = 5 \cdot KM^2$

Therefore:
 * $BK^2 : KM^2 = 5 : 1$

Therefore from :
 * $BK^2 : N^2 = 5 : 4$

which is not the ratio that a square number has to a square number.

Therefore from :
 * $BK$ is incommensurable in length with $N$.

Therefore $BK^2$ is greater than $KM^2$ by the square on a straight line which is incommensurable in length with $BK$.

We also have that $BK$ is commensurable with $BH$.

Therefore by :
 * $MB$ is a fourth apotome.

But the rectangle contained by a rational straight line and a fourth apotome is irrational.

By definition, its square root is irrational, and called minor.

Suppose $AH$ were joined.

Then $\triangle ABH$ is equiangular with $\triangle ABM$.

So:
 * $HB : BA = AB : BM$

Therefore:
 * $AB^2 = HB \cdot BM$

Therefore the side $AB$ of the pentagon $ABCDE$ is the irrational straight line called minor.