"vi. The angle in a semicircle is a right angle. It is believed that Thales proved this proposition in the following manner: Let ABCH be a circle of which the diameter is BC, and the centre E. ...Draw AE and produce BA to F. Because BE is equal to EA [both being radii of the circle], the angle EAB is equal to EBA; also, because AE is equal to EC, the angle EAC is equal to ECA [being angles at the base of an isosceles triangle]; wherefore, the whole angle BAC is equal to the two angles ABC, ACB. But FAC, the exterior angle of the triangle ABC, is also equal to the two angles ABC, ACB [since the sum of the three angles of the triangle is equal to two right angles, i.e., a straight line]; therefore the angle BAC is equal to the angle FAC, and each of them is therefore a right angle; wherefore the angle BAC in a semicircle is a right angle. Thales's demonstration, if we may call this his, is quite different from the one given in modern text-books; but it is certainly neither less rigid nor less beautiful. The demonstration is the one given in Euclid, but his work, we must remember, is to a large extent compiled from the works of previous writers. It will be seen, however, that this demonstration implies a knowledge of a seventh proposition,—"If one side of a triangle be produced, the exterior angle is equal to the sum of the two interior and opposite angles." Thales must have been familiar with this truth."