Let the two planes BA, BC be each of them perpendicular to a third plane, and let BD be the common section of the planes BA, BC: BD shall be perpendicular to the third plane.
For, if not, from the point D, draw in the plane BA, the straight line DE at right angles to AD, the common section of the plane BA with the third plane; [I. 11.
and from the point D, draw in the plane BC, the straight line DF at right angles to CD, the common section of the plane BC with the third plane. [I. 11.
Then, because the plane BA is perpendicular to the third plane, [Hypothesis.
and DE is drawn in the plane BA at right angles to AD their common section; [Construction.
therefore DE is perpendicular to the third plane. [XI. Def. 4.
In the same manner it may be shewn that DE is perpendicular to the third plane.
Therefore from the point D two straight lines are at right angles to the third plane, on the same side of it; which is impossible. [XI. 13.
Therefore from the point D, there cannot be any straight line at right angles to the third plane, except BD the common section of the planes BA, BC;
therefore BD is perpendicular to the third plane.
Wherefore, if two planes &c. q.e.d.
PROPOSITION 20. THEOREM.
If a solid angle be contained by three plane angles, any two of them are together greater than the third.
Let the solid angle at A be contained by the three plane angles BAC, CAD, DAB: any two of them shall be together greater than the third.
If the angles BAC, CAD, DAB be all equal, it is evident that any two of them are greater than the third.
