Page:The Construction of the Wonderful Canon of Logarithms.djvu/65

CONSTRUCTION OF THE CANON. 41 is the sine of the arc a d e. Draw a e; its half, f e, is the sine of the arc d e, the half of the arc a d e. Draw e c; its half, e g, is the sine of the arc e h, and is therefore the sine of the complement of the are de. Finally, make a k half the radius a b. Then as a k is to e f, so is e g to e i. For the two triangles c e a and c i e are equi-angular, since i c e or a c e is common to both; and c i e and c e a are each a right angle, the former by hypothesis, the latter because it is in the circumference and occupies a semicircle. Hence a c, the hypotenuse of the triangle c e a, is to a e, its less side, as e c, the hypotenuse of the triangle c i e, is to e i its less side. And since a c, the whole, is to a e as e c, the whole, is to e i, it follows that a b, half of a c, is to a e as e g, half of e c,is to e i. And now, finally, since a b, the whole, is to a e, the whole, as e g is to e i, we necessarily conclude that a k, half of a b, is to f e, half of a e, as e g is to e i.

Referring to the preceding figure, let the case be such that a e and e c are equal. In that case i will fall on b, and e i will be radius; also e f and e g will be equal, each of them being the sine of 45 degrees. Now (by 55) the ratio a k, half radius, to e f, a sine of 45 degrees, is likewise the ratio of e g, also a sine of 45