Exercises.
1. If three planes have a common line of intersection, the normals drawn to these planes from any point of that line are coplanar.
2. If two intersecting planes be respectively perpendicular to two intersecting lines, the line of intersection of the former is normal to the plane of the latter.
3. In the last case, show that the dihedral angle between the planes is equal to the rectilineal angle between the normals.
PROP. XX.—Theorem.
The sum of any two plane angles (BAD, DAC) of a trihedral angle (A) is greater than the third (BAC).
Dem.—If the third angle BAC be less than or equal to either of the other angles the proposition is evident. If not, suppose it greater: take any point D in AD, and at the point A in the plane BAC make the angle BAE equal BAD [I. xxiii.], and cut off AE equal AD. Through E draw BC, cutting AB, AC in the points B, C. Join DB, DC.
Then the triangles BAD, BAE have the two sides BA, AD in one equal respectively to the two sides BA, AE in the other, and the angle BAD equal to BAE; therefore the third side BD is equal to BE. But the sum of the sides BD, DC is greater than BC; hence DC is greater than EC. Again, because the triangles DAC, EAC have the sides DA, AC respectively equal to the sides EA, AC in the other, but the base DC greater than EC; therefore [I. xxv.] the angle DAC is greater than EAC, but the angle DAB is equal to BAE (const.). Hence the sum of the angles BAD, DAC is greater than the angle BAC.
PROP. XXI.—Theorem.