# 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.