:450px 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 {{EuclidPropLink|book = I|prop = 32|title = Sum of Angles of Triangle equals Two Right Angles}}: :$\angle ACL = \angle MFA$ 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 {{EuclidPropLink|book = V|prop = 18|title = Magnitudes Proportional Separated are Proportional Compounded}}: :$DC + CM : CM = MK : KF$ Therefore also: :$\left({DC + CM}\right)^2 : CM^2 = MK^2 : FK^2$ From {{EuclidPropLink|book = XIII|prop = 8|title = Straight Lines Subtending Two Consecutive Angles in Regular Pentagon cut in Extreme and Mean Ratio}}: :if $AC$ is cut in extreme and mean ratio, the greater segment equals $CD$. We have that: :$CM = \dfrac {AC} 2$ and so from {{EuclidPropLink|book = XIII|prop = 1|title = Area of Square on Greater Segment of Straight Line cut in Extreme and Mean Ratio}}: :$\left({DC + CM}\right)^2 = 5 \cdot CM^2$ 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 {{EuclidPropLink|book = X|prop = 9|title = Commensurability of Squares}}: :$BK$ 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 $MB$. {{qed|lemma}} 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 {{EuclidPropLink|book = X|prop = 15|title = Commensurability of Sum of Commensurable Magnitudes}}: :$KB$ is commensurable with $FB$. But we have that $BF$ is commensurable with $BH$. Therefore from {{EuclidPropLink|book = X|prop = 12|title = Commensurability is Transitive Relation}}: :$BK$ is commensurable with $BH$. We have that: :$BK^2 = 5 \cdot KM^2$ Therefore: :$BK^2 : KM^2 = 5 : 1$ Therefore from  {{EuclidPropLink|book = X|prop = 19|qual=Porism|title = Proportional Magnitudes have Proportional Remainders}}: :$BK^2 : N^2 = 5 : 4$ which is not the ratio that a square number has to a square number. Therefore from {{EuclidPropLink|book = X|prop = 9|title = Commensurability of Squares}}: :$BK$ 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 {{EuclidDefLink|X (III)|4|Fourth Apotome}}: :$MB$ is a fourth apotome. {{qed|lemma}} 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. {{qed}} {{Euclid Note|11|XIII}}