Proof: By Euclid

Let the given plane be the reference (plane), and $A$ a point in it.
So, it is required to set up a straight line at "right angles" to the reference plane at point $A$.

fig12e

Let some raised point $B$ have been assumed, and let the perpendicular (straight line) $BC$ have been drawn from $B$ to the reference plane [Prop. 11.11].
And let $AD$ have been drawn from point $A$ parallel to $BC$ [Prop. 1.31].
Therefore, since $AD$ and $CB$ are two parallel straight lines, and one of them, $BC$, is at "right angles" to the reference plane, the remaining (one) $AD$ is thus also at "right angles" to the reference plane [Prop. 11.8].
Thus, $AD$ has been set up at right angles to the given plane, from the point in it, $A$.
(Which is) the very thing it was required to do.
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