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

Therefore the square on HK is quadruple of the square on KC; therefore the squares on HK, KC, that is, the square on HC, is five times the square on KC.

But HC is equal to CB; therefore the square on BC is five times the square on CK.

And, since AB is double of CB, and, in them, AD is double of DB, therefore the remainder BD is double of the remainder DC.

Therefore BC is triple of CD; therefore the square on BC is nine times the square on CD.

But the square on BC is five times the square on CK; therefore the square on CK is greater than the square on CD; therefore CK is greater than CD.

Let CL be made equal to CK, from L let LM be drawn at right angles to AB, and let MB be joined.

Now, since the square on BC is five times the square on CK, and AB is double of BC, and KL double of CK, therefore the square on AB is five times the square on KL.

But the square on the diameter of the sphere is also five times the square on the radius of the circle from which the icosahedron has been described. [XIII. 16, Por.]

And AB is the diameter of the sphere; therefore KL is the radius of the circle from which the icosahedron has been described; therefore KL is a side of the hexagon in the said circle. [IV. 15, Por.]

And, since the diameter of the sphere is made up of the side of the hexagon and two of the sides of the decagon inscribed in the same circle, [XIII. 16, Por.] and AB is the diameter of the sphere, while KL is a side of the hexagon, and AK is equal to LB, therefore each of the straight lines AK, LB is a side of the decagon inscribed in the circle from which the icosahedron has been described.