PROPOSITION 21. THEOREM.
Every solid angle is contained by plane angles, which are together less than four right angles.
First let the solid angle at A be contained by three plane angles BAC, CAD, DAB: these three shall be together less than four right angles.
In the straight lines AB,AC,AD take any points B, C,D, and join BC, CD, DB.
Then, because the solid angle at B is contained by the three plane angles CBA, ABD, DBC, any two of them are together greater than the third, [XI. 20.
therefore the angles CBA, ABD are together greater than the angle DBC.
For the same reason, the angles BCA, ACD are together greater than the angle DCB,
and the angles CDA, ADB are together greater than the angle BDC.
Therefore the six angles CBA, ABD, BCA, ACD, CDA, ADB are together greater than the three angles DBC, DCB, BDC;
but the three angles DBC, DCB, BDC are together equal to two right angles. [I. 32.
Therefore the six angles CBA, ABD, BCA, ACD, CDA, ADB are together greater than two right angles.
And, because the three angles of each of the triangles ABC, ACD, ADB are together equal to two right angles, [1. 32.
therefore the nine angles of these triangles, namely, the angles CBA, BAC, ACB, ACD,CDA, CAD, ADD, DBA, DAB are equal to six right angles;
and of these, the six angles CBA, ACB, ACD, CDA, ADB, DBA are greater than two right angles,
therefore the remaining three angles BAC, CAD, DAB, which contain the solid angle at A, are together less than four right angles.
