Proof: By Euclid

For, if possible, let the two straight lines $AB$ and $AC$ have been set up at the same point $A$ at right angles to the reference plane, on the same side.
And let the plane through $BA$ and $AC$ have been drawn.

fig13e

So it will make a straight cutting (passing) through (point) $A$ in the reference plane [Prop. 11.3].
Let it have made $DAE$.
Thus, $AB$, $AC$, and $DAE$ are straight lines in one plane.
And since $CA$ is at right angles to the reference plane, it will thus also make right angles with all of the straight lines joined to it which are also in the reference plane [Def. 11.3] .
And $DAE$, which is in the reference plane, is joined to it.
Thus, angle $CAE$ is a right angle.
So, for the same (reasons), $BAE$ is also a right angle.
Thus, $CAE$ (is) equal to $BAE$.
And they are in one plane.
The very thing is impossible.
Thus, two (different) straight lines cannot be set up at the same point at right angles to the same plane, on the same side.
(Which is) the very thing it was required to show.
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