Proof: By Euclid
(related to Proposition: Prop. 11.11: Construction of Straight Line Perpendicular to Plane from point not on Plane)
 Let $A$ be the given raised point, and the given plane the reference (plane).

So, it is required to draw a perpendicular straight line from point $A$ to the reference plane.

Let some random straight line $BC$ have been drawn across in the reference plane, and let the (straight line) $AD$ have been drawn from point $A$ perpendicular to $BC$ [Prop. 1.12].
 If, therefore, $AD$ is also perpendicular to the reference plane then that which was prescribed will have occurred.
 And, if not, let $DE$ have been drawn in the reference plane from point $D$ at right angles to $BC$ [Prop. 1.11], and let the (straight line) $AF$ have been drawn from $A$ perpendicular to $DE$ [Prop. 1.12], and let $GH$ have been drawn through point $F$, parallel to $BC$ [Prop. 1.31].
 And since $BC$ is at right angles to each of $DA$ and $DE$, $BC$ is thus also at right angles to the plane through $EDA$ [Prop. 11.4].
 And $GH$ is parallel to it.
 And if two straight lines are parallel, and one of them is at right angles to some plane, then the remaining (straight line) will also be at right angles to the same plane [Prop. 11.8].
 Thus, $GH$ is also at right angles to the plane through $ED$ and $DA$.
 And $GH$ is thus at right angles to all of the straight lines joined to it which are also in the plane through $ED$ and $AD$ [Def. 11.3] .
 And $AF$, which is in the plane through $ED$ and $DA$, is joined to it.
 Thus, $GH$ is at right angles to $FA$.
 Hence, $FA$ is also at right angles to $HG$.
 And $AF$ is also at right angles to $DE$.
 Thus, $AF$ is at right angles to each of $GH$ and $DE$.
 And if a straight line is set up at right angles to two straight lines cutting one another, at the point of section, then it will also be at right angles to the plane through them [Prop. 11.4].
 Thus, $FA$ is at right angles to the plane through $ED$ and $GH$.
 And the plane through $ED$ and $GH$ is the reference (plane).
 Thus, $AF$ is at "right angles to the reference plane":bookofproofs$2212.
 Thus, the straight line $AF$ has been drawn from the given raised point $A$ perpendicular to the reference plane.
 (Which is) the very thing it was required to do.
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References
Adapted from (subject to copyright, with kind permission)
 Fitzpatrick, Richard: Euclid's "Elements of Geometry"