Perpendicular through Given Point

Theorem
Given an infinite straight line, and a given point not on that straight line, it is possible to draw a perpendicular to the given straight line.

Construction

 * Euclid-I-12.png

Let $AB$ be two points on the given infinite straight line.

Let $C$ be the given point not on it.

Let $D$ be some point not on $AB$ on the other side of it from $C$.

We construct a circle $EFG$ with center $C$ and radius $CD$.

We bisect the straight line $EG$ at the point $H$.

We draw line segments from $C$ to each of $G$, $H$ and $E$ to form the straight line segments $CG$, $CH$ and $CH$.

Then the line $CH$ is perpendicular to the given infinite straight line $AB$ through the given point $C$.

Proof
As $C$ is the center of circle $BCD$, it follows from that $GC = CE$.

As $EG$ has been bisected, $GH = HE$.

Thus, as $GC = CE$ and $GH = HE$, and $CH$ is common, by Triangle Side-Side-Side Equality‎, $\triangle CHG = \triangle EHG$.

Therefore $\angle CHG = \angle CHE$.

So $CH$ is a straight line set up on a straight line making the adjacent angles equal to one another.

Thus it follows from that each of $\angle CHG$ and $\angle CHE$ are right angles.

So the straight line $CH$ has been drawn at right angles to the given infinite straight line $AB$ through the given point $C$.

This theorem is one of the two attributed to Oenopides of Chios (the other being Proposition 23 of Book I).