Let a straight line AB be set up at right angles to the three straight lines BC, BD, and BE at their intersection B.
I say that BC, BD, and BE lie in one plane.
For suppose that they do not, but, if possible, let BD and BE lie in the plane of reference and BC in one more elevated. Produce the plane through AB and BC.
It intersects the plane of reference in a straight line. Let the intersection be BF. Therefore the three straight lines AB, BC, and BF lie in one plane, namely that drawn through AB and BC.
Now, since AB is at right angles to each of the straight lines BD and BE, therefore AB is also at right angles to the plane through BD and BE.
But the plane through BD and BE is the plane of reference, therefore AB is at right angles to the plane of reference.
Thus AB also makes right angles with all the straight lines which meet it and lie in the plane of reference.
But BF, which is the plane of reference, meets it, therefore the angle ABF is right. And, by hypothesis, the angle ABC is also right, therefore the angle ABF equals the angle ABC, and they lie in one plane, which is impossible.
Therefore the straight line BC is not in a more elevated plane. Therefore the three straight lines BC, BC, and BE lie in one plane.
Therefore, if a straight line is set up at right angles to three straight lines which meet one another at their common point of section, then the three straight lines lie in one plane.