Euclid, Elements
Thomas L. Heath, Sir Thomas Little Heath, Ed.

But, as BA is to AD, so is the square on BA to the square on AF, [V. Def. 9, VI. 8] for the triangle AFB is equiangular with the triangle AFD; therefore the square on BA is one and a half times the square on AF.

But the square on the diameter of the sphere is also one and a half times the square on the side of the pyramid. [XIII. 13]

And AB is the diameter of the sphere; therefore AF is equal to the side of the pyramid.

Again, since AD is double of DB, therefore AB is triple of BD.

But, as AB is to BD, so is the square on AB to the square on BF; [VI. 8, V. Def. 9] therefore the square on AB is triple of the square on BF.

But the square on the diameter of the sphere is also triple of the square on the side of the cube. [XIII. 15]

And AB is the diameter of the sphere; therefore BF is the side of the cube.

And, since AC is equal to CB, therefore AB is double of BC.

But, as AB is to BC, so is the square on AB to the square on BE; therefore the square on AB is double of the square on BE.

But the square on the diameter of the sphere is also double of the square on the side of the octahedron. [XIII. 14]

And AB is the diameter of the given sphere; therefore BE is the side of the octahedron.

Next, let AG be drawn from the point A at right angles to the straight line AB, let AG be made equal to AB, let GC be joined, and from H let HK be drawn perpendicular to AB.

Then, since GA is double of AC, for GA is equal to AB, and, as GA is to AC, so is HK to KC, therefore HK is also double of KC.