# Proposition 21

Any solid angle is contained by plane angles whose sum is less than four right angles.

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

I say that the sum of the angles BAC, CAD, and DAB is less than four right angles. Take points B, C, and D at random on the straight lines AB, AC, and AD respectively, and join BC, CD, and DB.

XI.20

Now, since the solid angle at B is contained by the three plane angles CBA, ABD, and CBD, and the sum of any two is greater than the remaining one, therefore the sum of the angles CBA and ABD is greater than the angle CBD.

For the same reason the sum of the angles BCA and ACD is greater than the angle BCD, and the sum of the angles CDA and ADB is greater than the angle CDB. Therefore the sum of the six angles CBA, ABD, BCA, ACD, CDA, and ADB is greater than the sum of the three angles CBD, BCD, and CDB.

I.32

But the sum of the three angles CBD, BDC, and BCD equals two right angles, therefore the sum of the six angles CBA, ABD, BCA, ACD, CDA, and ADB is greater than two right angles.

And, since the each sum of the three angles of the triangles ABC, ACD, and ADB equals two right angles, therefore the sum of the nine angles of the three triangles, the angles CBA, ACB, BAC, ACD, CDA, CAD, ADB, DBA, and BAD equals six right angles. Of them the sum of the six angles ABC, BCA, ACD, CDA, ADB, and DBA are greater than two right angles, therefore the sum of the remaining three angles BAC, CAD, and DAB containing the solid angle is less than four right angles.

Therefore, any solid angle is contained by plane angles whose sum is less than four right angles.

Q.E.D.

## Guide

In proposition XI.23 the condition stated here and the condition in XI.20 (the sum of any two plane angles is less than the third) together are shown to be sufficient to construct a solid angle.