Perpendicular through Given Point

Theorem

In the words of Euclid:

To a given infinite straight line, from a given point not on it, to draw a perpendicular straight line.

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 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 $C$.

As $C$ is the center of circle $BCD$, it follows from Book $\text{I}$ Definition $15$: Circle 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 Congruence, $\triangle CHG = \triangle EHG$.

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 Book $\text{I}$ Definition $10$: Right Angle 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 $C$.

$\blacksquare$