:450px Let $AB$ be the diameter of the given sphere. Let a regular dodecahedron and a regular icosahedron be inscribed within. Let $CDEFG$ be one pentagonal face of the dodecahedron. Let $KLH$ be one triangular face of the icosahedron. It is to be demonstrated that the radii of the circles circumscribing $CDEFG$ and $KLH$ are equal. Let $DG$ be joined. From {{EuclidPropLink|book = XIII|prop = 17|title = Construction of Regular Dodecahedron within Given Sphere}}: :$DG$ is the side of a cube which has been inscribed within the given sphere. Let $MN$ be a straight line such that: :$AB^2 = 5 \cdot MN^2$ From {{EuclidPropLink|book = XIII|prop = 16|qual=Porism|title = Construction of Regular Icosahedron within Given Sphere}}: :the square on the diameter of the given sphere is $5$ times the square on the radius of the circle in which is inscribed the pentagonal base of the pyramidal tip of a regular icosahedron which is inscribed within that sphere. Therefore $MN$ equals the radius of the circle in which that pentagon is inscribed. Let $MN$ be cut in extreme and mean ratio at $O$ such that $MO$ is the greater segment. Then from: :{{EuclidPropLink|book = XIII|prop = 9|title = Sides Appended of Hexagon and Decagon inscribed in same Circle are cut in Extreme and Mean Ratio}} and the converse of: :{{EuclidPropLink|book = XIII|prop = 5|title = Straight Line cut in Extreme and Mean Ratio plus its Greater Segment}} it follows that: :$MO$ is the side of a regular decagon which has been inscribed in the circle whose radius is $MN$. From {{EuclidPropLink|book = XIII|prop = 15|title = Construction of Cube within Given Sphere}}: :$5 \cdot MN^2 = AB^2 = 3 \cdot DG^2$ Let $DG$ be cut in extreme and mean ratio such that $CG$ is the greater segment. From the {{EuclidPropLink|book = XIV|prop = 8|qual=Lemma|title = Ratio of Volumes of Regular Dodecahedron and Regular Icosahedron in Same Sphere}}: :if two straight lines have been cut in extreme and mean ratio, their segments are in the same ratio. So: :$3 \cdot DG^2 : 3 \cdot CG^2 = 5 \cdot MN^2 : 5 \cdot MO^2$ From {{EuclidPropLink|book = XIII|prop = 16|title = Construction of Regular Icosahedron within Given Sphere}}: :$MO$ is the side of a regular decagon which has been inscribed and :$KL$ is the side of a regular pentagon which has been inscribed in the circle whose radius is $MN$. From {{EuclidPropLink|book = IV|prop = 15|qual=Porism|title = Inscribing Regular Hexagon in Circle}} :$MN$ is the side of a regular hexagon which has also been inscribed in that same circle. Thus from {{EuclidPropLink|book = XIII|prop = 10|title = Square on Side of Regular Pentagon inscribed in Circle equals Squares on Sides of Hexagon and Decagon inscribed in same Circle}}: :$5 \cdot MO^2 + 5 \cdot MN^2 = 5 \cdot KL^2$ Therefore: :$5 \cdot KL^2 = 3 \cdot CG^2 + 3 \cdot DG^2$ But from {{EuclidPropLink|book = XIII|prop = 12|title = Square on Side of Equilateral Triangle inscribed in Circle is Triple Square on Radius of Circle}}: :$5 \cdot KL^2$ equals the square on $15$ times the radius of the circle around $KLH$. From {{EuclidPropLink|book = XIV|prop = 2|qual=Lemma|title = Circle Circumscribing Pentagon of Dodecahedron and Triangle of Icosahedron in Same Sphere}}: :$3 \cdot DG^2 + 3 \cdot CG^2$ equals the square on $15$ times the radius of the circle around $CDEFG$. Therefore the two radii are equal. {{qed}} {{Euclid Note|2|XIV|This was proved by {{AuthorRef|Aristaeus the Elder}}.}}