Let any plane DE pass through AB and let CE be the common section of the planes DE, CK; [XI 3.
take any point F in CE, from which draw FG, in the plane DE, at right angles to CE. [I. 11.
Then, because AB is, at right angles to the plane CK, [Hypothesis. therefore it makes right angles with every straight line meeting it in that plane; [XI. Definition 3. but CB meets it, and is in that plane; therefore the angle ABF is a right angle. But the angle GFB is also a right angle; [Construction, therefore FG is parallel to AB. [I. 28. And AB is at right angles to the plane CK; [Hypothesis. therefore FG is also at right angles to the same plane. [XI. 8.
But one plane is at right angles to another plane, when the straight lines drawn in one of the planes at right angles to their common section, are also at right angles to the other plane; [XI. Definition 4.
and it has been shewn that any straight line FG drawn in the plane DE, at right angles to CE, the common section of the planes, is at right angles to the other plane CK;
therefore the plane DE is at right angles to the plane CK.
In the same manner it may be shewn that any other plane which passes through AB is at right angles to the plane CK
Wherefore, if a straight line &c., q.e.d.
PROPOSITION 19. THEOREM.
If two planes which cut one another be each of them perpendicular to a third plane, their common section shall be perpendicular to the same plane.
