Proof.

Let AB be a given finite straight line. Let the equilateral triangle ACB be constructed on it (proposition I.1), and let the angle ACB be bisected by the straight line CD (proposition I.9). Now AC is equal to BC and CD is common, and the angle ACD is equal to the angle BCD, hence AD is equal to BD (proposition I.4 (SAS)).

You can drag A and B.