If two parallel lines (AB, CD) be cut by three parallel planes (GH, KL, MN) in two triads of points (A, E, B; C, F, D), their segments between those points are proportional. Dem.—Join AC, BD, AD. Let AD meet the plane KL in X. Join EX, XF. Then because the parallel planes KL, MN are cut by the plane ABD in the lines EX, BD, these lines are parallel [XI. xvi.]. Hence In like manner, Therefore [V. xi.] PROP. XVIII.—Theorem. If a right line (AB) be normal to a plane (CI), any plane (DE) passing through it shall be perpendicular to that plane. Dem.—Let CE be the common section of the planes DE, CI. From any point F in CE draw FG in the plane DE parallel to AB [I. xxxi.]. Then because AB and FG are parallel, but AB is normal, to the plane CI; hence FG is normal to it [XI. viii.]. Now since FG is parallel to AB, the angles ABF, BFG are equal to two right angles [I. xxix.]; but ABF is right (hyp.); therefore BFG is right—that is, FG is perpendicular to CE. Hence every line in the plane DE, drawn perpendicular to the common section of the planes DE, CI, is normal to the plane CI. Therefore [XI. Def. viii.] the planes DE, CI are perpendicular to each other. PROP. XIX.—Theorem. If two intersecting planes (AB, BC) be each perpendicular to a third plane (ADC), their common section (BD) shall be normal to that plane. Dem.—If not, draw from D in the plane AB the line DE perpendicular to AD, the common section of the planes AB, ADC; and in the plane BC draw BF perpendicular to the common section DC of the planes BC, ADC. Then because the plane AB is perpendicular to ADC, the line DE in AB is normal to the plane ADC [XI. Def. viii.]. In like manner DF is normal to it. Therefore from the point D there are two distinct normals to the plane ADC, which [XI. xiii.] is impossible. Hence BD must be normal to the plane ADC. 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.