Three Intersecting Lines Perpendicular to Another Line are in One Plane

Proof

 * Euclid-XI-5.png

Let $AB$ be a straight line set up at right angles to the three straight lines $BC$, $BD$ and $BE$ at their point of meeting $B$.

It is to be demonstrated that $BC$, $BD$ and $BE$ are all in the same plane.

Suppose they are not all in the same plane, but that while $BD$ and $BE$ are in the plane of reference, $BC$ is in a plane more elevated.

Thus the plane through $AB$ and $BC$ is different from the plane of reference.

From :
 * the common section of the plane through $AB$ and $BC$ with the plane of reference is a straight line $BF$, say.

Therefore the three straight lines $AB$, $BC$ and $BF$ are in the same plane, that is, the one through $AB$ and $BC$.

We have that $AB$ is at right angles to each of the straight lines $BD$ and $BE$.

Therefore from :
 * $AB$ is 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 from :
 * $AB$ will also make right angles with all the straight lines which meet it and are in the plane of reference.

But $BF$, which is in the plane of reference, meets $AB$.

Therefore $\angle ABF$ is a right angle.

But by hypothesis $\angle ABC$ is also a right angle.

Therefore $\angle ABF = \angle ABC$.

But $\angle ABF$ and $\angle ABC$ are in one plane, which is impossible.

Therefore $BC$ is not in a plane more elevated.

Therefore $BC$, $BD$ and $BE$ are all in the same plane.