Proof: (Euclid)

Let the point $D$ be have been taken at random on $AC$, and let $CE$ be made equal to $CD$ [Prop. 1.3], and let the equilateral triangle $FDE$ have been constructed on $DE$ [Prop. 1.1], and let $FC$ have been joined.
I say that the straight line $FC$ has been drawn at right angles to the given straight line $AB$ from the given point $C$ on it.
For since $DC$ is equal to $CE$, and $CF$ is common, the two (straight lines) $DC$, $CF$ are equal to the two (straight lines), $EC$, $CF$, respectively.
And the base $DF$ is equal to the base $FE$.
Thus, the angle $DCF$ is equal to the angle $ECF$ [Prop. 1.8], and they are adjacent.
But when a straight line stood on a(nother) straight line makes the adjacent angles equal to one another, each of the equal angles is a [Def. 1.10] .
Thus, each of the (angles) $DCF$ and $FCE$ is a right angle.
Thus, the straight line $CF$ has been drawn at right angles to the given straight line $AB$ from the given point $C$ on it.
(Which is) the very thing it was required to do.
∎

Thank you to the contributors under CC BY-SA 4.0!