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

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

Let a point D be taken at random on AC;
let CE be made equal to CD; [I. 3]
on DE let the equilateral triangle FDE be constructed,
and let FC be joined;
I say that the straight line FC has been drawn at right angles to the given straight line AB from C the given point on it.

For, since DC is equal to CE,
and CF is common,
the two sides DC, CF are equal to the two sides EC, CF respectively;
and the base DF is equal to the base FE;
therefore the angle DCF is equal to the angle ECF; [I. 8]
and they are adjacent angles.

But, when a straight line set up on a straight line makes the adjacent angles equal to one another, each of the equal angles is right; [Def. 10]
therefore each of the angles DCF, 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.