Proposition 13
Set out the diameter AB of the given sphere, cut it at the point C so that AC is double CB, describe the semicircle ADB on AB, draw CD from the point C at right angles to AB, and join DA.
Set out the circle EFG with radius equal to DC, inscribe the equilateral triangle EFG in the circle EFG, take the center H of the circle, and join EH, HF, and HG.
Set HK up from the point H at right angles to the plane of the circle EFG, cut off HK equal to the straight line AC from HK, and join KE, KF, and KG.
Now, since KH is at right angles to the plane of the circle EFG, therefore it makes right angles with all the straight lines which meet it and are in the plane of the circle EFG. But each of the straight lines HE, HF, and HG meets it, therefore HK is at right angles to each of the straight lines HE, HF, and HG.
And, since AC equals HK, and CD equals HE, and they contain right angles, therefore the base DA equals the base KE. For the same reason each of the straight lines KF and KG also equals DA. Therefore the three straight lines KE, KF, and KG equal one another.