Construction of Perpendicular Line

Construction

 * Euclid-I-11.png

Let $AB$ be the given straight line segment, and let $C$ be the given point on it.

Let a point $D$ be taken on $AB$.

We cut off from $CB$ a length $CE$ equal to $DC$.

We construct an equilateral triangle $\triangle DEF$ on $DE$.

We draw the line segment $FC$.

Then $FC$ is the required perpendicular to $AB$.

Proof
Since $DC = CE$ and $FC$ is common to both, and $DF = FE$, triangle $\triangle DCF$ equals triangle $\triangle ECF$.

Thus $\angle DCF = \angle ECF$.

So $CF$ 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 $\angle DCF$ and $\angle ECF$ are right angles.

So the straight line $CF$ has been drawn at right angles to the given straight line $AB$ from the given point $C$ on it.