The First Six Books of the Elements of Euclid

by Euclid; John Casey

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.