5. If the intersections of several planes be parallel, the normals drawn to them from any point are coplanar.
PROP. XV.—Theorem.
Two planes (AC, DF) are parallel, if two intersecting lines (AB, BC) on one of them be respectively parallel to two intersecting lines (DE, EF) on the other.
Dem.—From B draw BG perpendicular to the plane DF [XI. xi.], and let it meet that plane in G. Through G draw GH parallel to ED, and GK to EF. Now, since GH is parallel to ED (const.), and AB to ED (hyp.), AB is parallel to GH [XI. ix.]. Hence the sum of the angles ABG, BGH is two right angles [I. xxix]; but BGH is right (const.); therefore ABG is right. In like manner CBG is right. Hence BG is normal to the plane AC [XI. Def. vi.], and it is normal to DF (const.). Hence the planes AC, DF have a common normal BG; therefore they are parallel to one another.
PROP. XVI.—Theorem.
If two parallel planes (AB, CD) be cut by a third plane (EF, HG), their common sections (EF, GH) with it are parallel.
Dem.—If the lines EF, GH are not parallel, they must meet at some finite distance. Let them meet in K. Now since K is a point in the line EF, and EF is in the plane AB, K is in the plane AB. In like manner K is a point in the plane CD. Hence the planes AB, CD meet in K, which is impossible, since they are parallel. Therefore the lines EF, GH must be parallel.
Exercises.