 (diagram 1) Let the given straight-line be AB, the given point on it G. It is, in fact, required to draw a straight line from point G at right angles to straight-line AB. (diagram 2) Let there be taken on AG a point as happens, D, (diagram 3) and let an equal, GE, to GD be positioned, (diagram 4) let an equilateral triangle, ZDE, be constructed on DE, (diagram 5) and let ZG be joined. I say that a straight line, ZG, has been drawn at right angles to the given straight-line AB from the given point on it, G. For since DG is equal to G and GZ is common, in fact, two, DG, GZ, are respectively equal to two, EG, GZ. And a base, DZ, is equal to a base, ZE. (diagram 6) Therefore, an angle, that by DGZ, is equal that by EGZ. And they are in succession. But whenever a straight-line stood on a straight-line makes successive angles equal to one another, each of the equal angles is right. Therefore, each of the angles by DGZ, ZGE, is right. Therefore, a straight line, GZ, has been drawn at right angles to the given straight-line, AB, from the given point on it, G, which it was require to make.