To draw a straight line at right angles to a given straight line from a given point on it.

Let *AB* be the given straight line, and *C* the given point on it.

It is required to draw a straight line at right angles to the straight line *AB* from the point *C.*

Take an arbitrary point *D* on *AC.* Make *CE* equal to *CD.* Construct the equilateral triangle *FDE* on *DE,* and join *CF.*

I say that the straight line *CF* has been drawn at right angles to the given straight line *AB* from *C* the given point on it.

Since *CD* equals *CE,* and *CF* is common, therefore the two sides *CD* and *CF* equal the two sides *CE* and *CF* respectively, and the base *DF* equals the base *EF.* Therefore the angle *DCF* equals the angle *ECF,* and they are adjacent angles.

But, when a straight line standing on a straight line makes the adjacent angles equal to one another, each of the equal angles is right, therefore each of the angles *DCF* and *FCE* is right.

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

Q.E.F.

## Construction stepsThe actual construction here is the same double-equilateral-triangle construction of the previous proposition that is used to bisect the lineDE, except that it is preceded by the selection of points D and E on AB equidistant from C.
This construction actually only requires drawing three circles and the one line |