:500px Let $AB$ be the diameter of the given sphere. Let $AB$ be cut at $C$ where $AC = 2 \cdot CB$. Let $ADB$ be a semicircle on the diameter $AB$. Let $CD$ be drawn from $C$ perpendicular to $AB$. Let $DA$ be joined. Let the circle $EFG$ be constructed whose radius equals $DC$. From {{EuclidPropLink|book = IV|prop = 2|title = Inscribing in Circle Triangle Equiangular with Given}}: :let the equilateral triangle $EFG$ be inscribed within the circle $EFG$. From {{EuclidPropLink|book = III|prop = 1|title = Finding Center of Circle}}: :let the center of the circle $EFG$ be $H$. Let $EH, HF, HG$ be joined. From {{EuclidPropLink|book = XI|prop = 12|title = Construction of Straight Line Perpendicular to Plane from point on Plane}}: :let $HK$ be drawn from $H$ perpendicular to the plane of the circle $EFG$. Let $HK$ equal to $AC$ be cut off from $HK$. Let $KE, KF, KG$ be joined. We have that $KH$ is perpendicular to the plane of the circle $EFG$. So from {{EuclidDefLink|XI|3|Line at Right Angles to Plane}}: :$KH$ is perpendicular to each of the straight lines which meet it and are in the plane of the circle $EFG$. But each of $HE, HF, HG$ join $HK$ and are in the plane of the circle $EFG$. Therefore $HK$ is perpendicular to each of $HE, HF, HG$. We have that: :$AC = HK$ and: :$CD = HE$ Both $AC, CD$ and $HK, HE$ contain right angles. Therefore from {{EuclidPropLink|book = I|prop = 4|title = Triangle Side-Angle-Side Equality}}: :$DA = KE$ For the same reason: :$KF = DA$ :$KG = DA$ Therefore the three straight lines $KE, KF$ and $KG$ are equal to one another. We have that: :$AC = 2 \cdot CB$ Therefore: :$AB = 3 \cdot CB$ But from {{EuclidPropLink|book = XIII|prop = 13|qual=Lemma|title = Construction of Regular Tetrahedron within Given Sphere}}: :$AB : BC = AD^2 : DC^2$ Therefore: :$AD^2 = 3 \cdot DC^2$ But from {{EuclidPropLink|book = XIII|prop = 12|title = Square on Side of Equilateral Triangle inscribed in Circle is Triple Square on Radius of Circle}}: :$FE^2 = 3 \cdot EH^2$ We have that: :$DC = EH$ Therefore: :$DA = EF$ But we have that: :$DA = KE = KF = KG$ Therefore: :$EF = FG = GE = KE = KF = KG$ Therefore each of $\triangle EFG, \triangle KEF, \triangle KFG, \triangle KEG$ are equilateral. Therefore a regular tetrahedron $EFGK$ has been constructed out of four equilateral triangles. In this construction, $\triangle EFG$ is its base and $K$ is its apex. {{qed|lemma}} It is next to be demonstrated that $EFGK$ is inscribed in the given sphere. Let $KH$ be produced to $L$ such that $HL = CB$. From {{EuclidPropLink|book = VI|prop = 8|qual=Porism|title = Perpendicular in Right-Angled Triangle makes two Similar Triangles}}: :$AC : CD = CD : CB$ while: :$AC = KH$ :$CD = HE$ :$CB = HL$ Therefore: :$KH : HE = EH : HL$ Therefore from {{EuclidPropLink|book = VI|prop = 17|title = Rectangles Contained by Three Proportional Straight Lines}}: :$KH \cdot HL = EH^2$ We have that $\angle KHE$ and $\angle EHL$ are both right angles. Therefore from: :{{EuclidPropLink|book = VI|prop = 8|title = Perpendicular in Right-Angled Triangle makes two Similar Triangles}} and: :{{EuclidPropLink|book = III|prop = 31|title = Relative Sizes of Angles in Segments}} the semicircle described on $KL$ also passes through $E$. Let $KL$ remain fixed, and let that semicircle be carried around and restored to the same position from which it began. Then it will pass through the points $F$ and $G$. We have that: :$KH = AC$ and: :$HL = AB$ Thus: :$KL = AB$ Therefore $KL$ is the diameter of the given sphere. Thus the regular tetrahedron $EFGK$ has been inscribed in the given sphere. {{qed|lemma}} It remains to be shown that the square on the diameter of the given sphere is $1 \frac 1 2$ the square on the side of the regular tetrahedron $EFGK$. We have that: :$AC = 2 \cdot CB$ Therefore: :$AB = 3 \cdot CB$ and so: :$2 \cdot AB = 3 \cdot AC$ But: :$BA : AC = BA^2 : AD^2$ Therefore: :$BA^2 = \dfrac 3 2 AD^2$ But we have that: :$BA$ equals the diameter of the given sphere and: :$AD$ equals the side of the regular tetrahedron $EFGK$. Hence the result. {{qed}} {{Euclid Note|13|XIII}}