If two planes which cut one another are at right angles to any plane, then their intersection is also at right angles to the same plane.

Let the two planes *AB* and *BC* be at right angles to the plane of reference, and let *BC* be their intersection.

I say that *BD* is at right angles to the plane of reference.

Suppose it is not. From the point *D* draw *DE* at right angles to the straight line *AD* in the plane *AB,* and draw *DF* at right angles to *CD* in the plane *BC.*

Now, since the plane *AB* is at right angles to the plane of reference, and *DE* is at right angles in the plane *AB* to *AD,* their intersection, therefore *DE* is at right angles to the plane of reference.

Similarly we can prove that *DF* is also at right angles to the plane of reference. Therefore from the same point *D* two straight lines have been set up at right angles to the plane of reference on the same side, which is impossible.

Therefore no straight line except the intersection *DB* of the planes *AB* and *BC* can be set up from the point *D* at right angles to the plane of reference.

Therefore, *if two planes which cut one another are at right angles to any plane, then their intersection is also at right angles to the same plane.*

Q.E.D.