2. If the point be above the plane, there can be but one normal; for, if there could be two, they would be parallel [XI. vi.] to one another, which is absurd. Therefore from the same point there can be drawn but one normal to a given plane.
PROP. XIV.—Theorem.
Planes (CD, EF) which have a common normal (AB) are parallel to each other.
Dem.—If the planes be not parallel, they will meet when produced. Let them meet, their common section being the line GH, in which take any point K. Join AK, BK. Then because AB is normal to the plane CD, it is perpendicular to the line AK, which it meets in that plane [XI. Def. vi.]. Therefore the angle BAK is right. In like manner the angle ABK is right. Hence the plane triangle ABK has two right angles, which is impossible. Therefore the planes CD, EF cannot meet—that is, they are parallel.
Exercises.
1. The angle between two planes is equal to the angle between two intersecting normals to these planes.
2. If a line be parallel to each of two planes, the sections which any plane passing through it makes with them are parallel.
3. If a line be parallel to each of two intersecting planes, it is parallel to their intersection.
4. If two right lines be parallel, they are parallel to the common section of any two planes passing through them.