:400px Let $ABCDE$ be a circle. Let the regular pentagon $ABCDE$ be inscribed within the circle $ABCDE$. It is to be demonstrated that the square on the side of the pentagon $ABCDE$ equals the square on the side of a regular hexagon plus the square on the side of a regular decagon which have both been inscribed within the circle $ABCDE$. Let the center of the circle $ABCDE$ be $F$. Let $AF$ be joined and produced to $G$. Let $FB$ be joined. Let $FH$ be drawn from $F$ perpendicular to $AB$ and produced to $K$. Let $AK$ and $KB$ be joined. Let $FL$ be drawn from $F$ perpendicular to $AK$ and produced to $M$. Let $KN$ be joined. We have that the arc $ABCG$ equals the arc $AEDG$. Within those arcs, the arc $ABC$ equals the arc $AED$. Therefore the remainders are equal: the arc $CG$ equals the arc $GD$. But $CD$ is the side of a regular pentagon. Therefore $CG$ (when joined) is the side of a regular decagon. We have that: :$FA = FB$ and: :$FH \perp AB$ Therefore from: :{{EuclidPropLink|book = I|prop = 5|title = Isosceles Triangle has Two Equal Angles}} and: :{{EuclidPropLink|book = I|prop = 26|title = Triangle Side-Angle-Angle Equality}}: it follows that: :$\angle AFK = \angle KFB$ Hence from {{EuclidPropLink|book = III|prop = 26|title = Equal Angles in Equal Circles}}: :the arc $AK$ equals the arc $KB$. Therefore: :the arc $AB$ equals double the arc $BK$. Therefore $AK$ is the side of a regular decagon. For the same reason: :the arc $AK$ equals double the arc $KM$. We have that: :the arc $AB$ equals double the arc $BK$ while: :the arc $CD$ equals the arc $KB$. Therefore: :the arc $CD$ equals double the arc $BK$. But: :the arc $AK$ equals double the arc $KM$. Therefore: :the arc $BK$ equals double the arc $KM$ and therefore: :the arc $CG$ equals double the arc $KM$. But we have that: :the arc $CB$ equals the arc $BA$ therefore :the arc $CB$ equals double the arc $BK$. Therefore: :the arc $GB$ equals double the arc $BM$. So from {{EuclidPropLink|book = VI|prop = 33|title = Angles in Circles have Same Ratio as Arcs}}: :$\angle GFB = 2 \cdot \angle BFM$ But we also have: :$\angle FAB = \angle ABF$ and so: :$\angle GFB = 2 \cdot \angle FAB$ Therefore: :$\angle BFN = \angle FAB$ But $\angle ABF$ is common between $\triangle ABF$ and $\triangle BFN$. Therefore from {{EuclidPropLink|book = I|prop = 32|title = Sum of Angles of Triangle equals Two Right Angles}}: :$\angle AFB = \angle BNF$ Therefore from {{EuclidPropLink|book = VI|prop = 4|title = Equiangular Triangles are Similar}}: :$AB : BF = FB : BN$ Therefore from {{EuclidPropLink|book = VI|prop = 17|title = Rectangles Contained by Three Proportional Straight Lines}}: :$AB \cdot BN = BF^2$ We have that: :$AL = LK$ while $LN$ is common and perpenducular to $AK$. Therefore from {{EuclidPropLink|book = I|prop = 4|title = Triangle Side-Angle-Side Equality}}: :$KN = AN$ Therefore: :$\angle LKN = \angle LAN$ But: :$\angle LAN = \angle KBN$ Therefore: :$\angle LAN = \angle KBN$ We have that: :$\angle A$ is common to $\triangle AKB$ and $\triangle AKN$. Therefore from {{EuclidPropLink|book = I|prop = 32|title = Sum of Angles of Triangle equals Two Right Angles}}: :$\angle AKB = \angle KNA$ Therefore: :$\triangle AKB$ is equiangular with $\triangle KNA$. Therefore from {{EuclidPropLink|book = VI|prop = 4|title = Equiangular Triangles are Similar}}: :$BA : AK = KA : AN$ Therefore from {{EuclidPropLink|book = VI|prop = 17|title = Rectangles Contained by Three Proportional Straight Lines}}: :$BA \cdot AN = AK^2$ But we have that: :$AB \cdot BN = BF^2$ Therefore: :$AB \cdot BN + BA \cdot AN = BF^2 + AK^2$ But from {{EuclidPropLink|book = II|prop = 2|title = Square is Sum of Two Rectangles}}: :$AB \cdot BN + BA \cdot AN = BA^2$ That is: :$BA^2 = BF^2 + AK^2$ But: : $BA$ is the side of the pentagon $ABCDE$ : $AK$ is the side of a regular decagon inscribed within the circle $ABCDE$. and from {{EuclidPropLink|book = IV|prop = 15|qual=Porism|title = Inscribing Regular Hexagon in Circle}}: : $BF$ is the side of a regular hexagon inscribed within the circle $ABCDE$. Hence the result. {{qed}} {{Euclid Note|10|XIII}}