Proof: By Euclid
(related to Proposition: Prop. 11.05: Three Intersecting Lines Perpendicular to Another Line are in One Plane)
 For let some straight line $AB$ have been set up at right angles to three straight lines $BC$, $BD$, and $BE$, at the (common) point of section $B$.

I say that $BC$, $BD$, and $BE$ are in one plane.

For (if) not, and if possible, let $BD$ and $BE$ be in the reference plane, and $BC$ in a more elevated (plane).
 And let the plane through $AB$ and $BC$ have been produced.
 So it will make a straight line as a common section with the reference plane [Def. 11.3] .
 Let it make $BF$.
 Thus, the three straight lines $AB$, $BC$, and $BF$ are in one plane  (namely), that drawn through $AB$ and $BC$.
 And since $AB$ is at right angles to each of $BD$ and $BE$, $AB$ is thus also at right angles to the plane (passing) through $BD$ and $BE$ [Prop. 11.4].
 And the plane (passing) through $BD$ and $BE$ is the reference plane.
 Thus, $AB$ is at "right angles to the reference plane":bookofproofs$2212.
 Hence, $AB$ will also make right angles with all straight lines joined to it which are also in the reference plane [Def. 11.3] .
 And $BF$, which is in the reference plane, is joined to it.
 Thus, the angle $ABF$ is a right angle.
 And $ABC$ was also assumed to be a right angle.
 Thus, angle $ABF$ (is) equal to $ABC$.
 And they are in one plane.
 The very thing is impossible.
 Thus, $BC$ is not in a more elevated plane.
 Thus, the three straight lines $BC$, $BD$, and $BE$ are in one plane.
 Thus, if a straight line is set up at right angles to three straight lines cutting one another, at the (common) point of section, then the three straight lines are in one plane.
 (Which is) the very thing it was required to show.
∎
Thank you to the contributors under CC BYSA 4.0!
 Github:

 nonGithub:
 @Fitzpatrick
References
Adapted from (subject to copyright, with kind permission)
 Fitzpatrick, Richard: Euclid's "Elements of Geometry"