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 Definition I-15 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 Definition I-10 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).