Construct the segments CE and CF.
Since both CE and CF are radii of the same circle they are equal.
Let AB be the given infinite straight line and let C be the given point.
Construct a circle D with center at C having radius larger than the distance from C to the line AB.
Label the points of intersection of the circle and the line to be E and F.
Construct the bisector CG of the angle ECF.
Then the triangles ECG and FCG are congruent. For the angles
ECG and FCG are equal, the sides EC and FC are
equal, and they share the common side CG.
(Prop I-4)
Therefore the angles EGC and FGC are equal.
Moreover, these angles are adjacent. Finally, because the angles are equal
they must be right angles.
(Defn-10)
Therefore CG is perpendicular to EF and thus to AB, which was to be proved.
QED
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