Next, let the solid angle at A be contained by any number of plane angles BAC, CAD, DAE, EAF, FAB: these shall be together less than four right angles.
Let the planes in which the angles are, be cut by a plane, and let the common sections of it with those planes be BC, CD, DE, EF, FB.
Then, because the solid angle at B is contained by the three plane angles CBA, ABF, FBC, any two of them are together greater than the third, [XL 20.
therefore the angles CBA, ABF are together greater than the angle FBC.
For the same reason, at each of the points C, D, E, F, the two plane angles which are at the bases of the triangles having the common vertex A, are together greater than the third angle at the same point, which is one of the angles of the polygon BCDEF.
Therefore all the angles at the bases of the triangles are together greater than all the angles of the polygon.
Now all the angles of the triangles are together equal to twice as many right angles as there are triangles, that is, as there are sides in the polygon BCDEF; [I. 32.
and all the angles of the polygon, together with four right angles, are also equal to twice as many right angles as there are sides in the polygon; [I. 32, Corollary 1.
therefore all the angles of the triangles are equal to all the angles of the polygon, together with four right angles. [Ax. 1.
But it has been shewn that all the angles at the bases of the triangles are together greater than all the angles of the polygon;
therefore the remaining angles of the triangles, namely, those at the vertex, which contain the solid angle at A, are together less than four right angles.
Wherefore, every solid angle &c. q.e.d.
