Construction of Perpendicular Line

Theorem
It is possible to draw a straight line at right angles to a given straight line from a given point on it.

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 Definition I-10 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.