PROPOSITION 11.¶
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;[1] [I. 3]
on DE let the equilateral triangle FDE be constructed, [I. 1]
and let FC be joined; [Post. 1]
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,
and the base DF is equal to the base FE;
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 the straight line CF has been drawn at right angles to the given straight line AB from the given point C on it.
See Commentary on Proposition I.11.