Note 25 — BOOK III. (continued 3)

This will be manifest from the following figure. Let the circles a c, b d, be drawn passing through their respective centres a, b; and from the centre c, with the radius c b, equal to a b, describe the arch a b d, and draw the lines c b, c d, c a. Then because a c b is an equilateral triangle, as also c b d, each of the angles a c b, b c d, shall be equal to ⅔ of one right angle; and because the biline c d, is equal to the biline c b, hence, the angle formed by the arch c b, and the arch c d, viz. the angle a c f, shall be equal to the angle formed by the right line c b, and the right line c d, i.e. to ⅔ of one right angle. Q. E. D.