The sum of all the plane angles (BAC, CAD, &c.) forming any solid angle (A) is less than four right angles.
Dem.—Suppose for simplicity that the solid angle A is contained by five plane angles BAC, CAD, DAE, EAF, FAB; and let the planes of these angles be cut by another plane in the lines BC, CD, DE, EF, FB; then we have [XI. xx.],
| ∠ABC + ABF | greater than | FBC, |
| ∠ACB + ACD | ,, | BCD, &c. |
Hence, adding, we get the sum of the base angles of the five triangles BAC, CAD, &c., greater than the sum of the interior angles of the pentagon BCDEF —that is, greater than six right angles. But the sum of the base angles of the same triangles, together with the sum of the plane angles BAC, CAD, &c., forming the solid angle A, is equal to twice as many right angles as there are triangles BAC, CAD, &c.—that is, equal to ten right angles. Hence the sum of the angles forming the solid angle is less than four right angles.
Observation.—This Prop. may not hold if the polygonal base BCDEF contain re-entrant angles.
Exercises on Book XI.
1. Any face angle of a trihedral angle is less than the sum, but greater than the difference, of the supplements of the other two face angles.
2. A solid angle cannot be formed of equal plane angles which are equal to the angles of a regular polygon of n sides, except in the case of n = 3, 4, or 5.
3. Through one of two non-coplanar lines draw a plane parallel to the other.
4. Draw a common perpendicular to two non-coplanar lines, and show that it is the shortest distance between them.