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

== Theorem ==
== Proof ==
:
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.
== Sources ==
* : Book $\text{XIII}$. Propositions
Category:Pentagons