Proposition 20

If a solid angle is contained by three plane angles, then the sum of any two is greater than the remaining one.

Let the solid angle at A be contained by the three plane angles BAC, CAD, and DAB.

I say that the sum of any two of the angles BAC, CAD, and DAB is greater than the remaining one.

If the angles BAC, CAD, and DAB are equal to one another, then it is clear that the sum of any two is greater than the remaining one.

But, if not, let BAC be greater. In the plane through BA and AC, construct the angle BAE equal to the angle DAB at the point A on the straight line AB. Make AE equal to AD, draw BEC across through the point E cutting the straight lines AB and AC at the points B and C, and join DB and DC.

I.4

Now, since DA equals AE, and AB is common, therefore two sides are equal to two sides. And the angle DAB equals the angle BAE, therefore the base DB equals the base BE.

I.20

And, since the sum of the two sides BD and DC is greater than BC, and of these DB was proved equal to BE, therefore the remainder DC is greater than the remainder EC.

I.25

Now, since DA equals AE, and AC is common, and the base DC is greater than the base EC, therefore the angle DAC is greater than the angle EAC.

But the angle BAE equals the angle DAB, therefore the sum of the angles DAB and DAC is greater than the angle BAC.

Similarly we can prove that the sum of any two of the remaining angles is greater than the remaining one.

Therefore, if a solid angle is contained by three plane angles, then the sum of any two is greater than the remaining one.

Q.E.D.

Guide

This is one of two necessary conditions for constructing a solid angle out of three plane angles. The next necessary condition is stated in the next proposition, and the two conditions together are shown to be sufficient in XI.23.

About the proof

The structure of the proof is not entirely clear. The goal is to show that the sum of any two of the angles is greater than the third. Notice is made that if they are all equal, then the goal is clearly satisfied. Then, under the assumption that if one is greater than a second, then the sum of the second and third is greater than the first. Then the other cases are declared to be similarly provable.

Various interpretations have been made of the intent of the form of proof, but all require minor changes to clarify the structure.

About three-dimensional analogues of two-dimensional constructions

Up until this proposition, each construction in Book XI takes place within a plane, although different constructions in the same proposition may occur in different planes. One of the constructions here, however, takes place in two different planes. The angle BAE is constructed in one plane to equal a given angle DAB in a different plane. The construction in I.23 to construct one angle equal to a given angle, strictly speaking, takes place in only one plane. Tracing that construction back through Book I leads through proposition I.22 to I.3. Proposition I.3 cuts one line off equal to another line. That basic construction can easily be modified so that the two lines are in different planes. Once that’s done, the rest of the constructions in Book I also apply when their components lie in different planes. Still, the details should be verified before applying the results as done in the proof of this proposition.