previous next


If then, KL remaining fixed, the semicircle be carried round and restored to the same position from which it began to be moved, it will also pass through the points F, G, since, if FL, LG be joined, the angles at F, G similarly become right angles; and the pyramid will be comprehended in the given sphere.

For KL, the diameter of the sphere, is equal to the diameter AB of the given sphere, inasmuch as KH was made equal to AC, and HL to CB.

I say next that the square on the diameter of the sphere is one and a half times the square on the side of the pyramid

For, since AC is double of CB, therefore AB is triple of BC; and, convertendo, BA is one and a half times AC.

But, as BA is to AC, so is the square on BA to the square on AD.

Therefore the square on BA is also one and a half times the square on AD.

And BA is the diameter of the given sphere, and AD is equal to the side of the pyramid.

Therefore the square on the diameter of the sphere is one and a half times the square on the side of the pyramid. Q. E. D.


LEMMA.

It is to be proved that, as AB is to BC, so is the square on AD to the square on DC.

For let the figure of the semicircle be set out, let DB be joined, let the square EC be described on AC, and let the parallelogram FB be completed.

Since then, because the triangle DAB is equiangular with the triangle DAC, as BA is to AD, so is DA to AC, [VI. 8, VI. 4]

Creative Commons License
This work is licensed under a Creative Commons Attribution-ShareAlike 3.0 United States License.

An XML version of this text is available for download, with the additional restriction that you offer Perseus any modifications you make. Perseus provides credit for all accepted changes, storing new additions in a versioning system.

hide Display Preferences
Greek Display:
Arabic Display:
View by Default:
Browse Bar: