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.*

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.*

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.

This proposition is used in the proof of the remark after proposition XIII.18 to show that the five regular polyhedra constructed in Book XIII are the only five possible.