Proof.

Let AB be a given straight line and C a point on it. Take point D at random on AC, and make CE equal to DC (proposition I.3).

On DE, construct the equilateral triangle DEF (proposition I.1). Draw CF.

Since DC is equal to CE, CF is common, and the two sides DF and EF are equal, triangles DCF and ECF are congruent (SSS proposition I.8). In particular, the angle ECF is equal to the angle DCF and hence, per definition, these angles are right angles and the line CF is perpendicular to AB.

Drag e.g. A, B, C and D.