Page:Journal of the Asiatic Society of Bengal Vol 1.djvu/556

502 .-. B D = B H, and it has been proved to be greater, which is ab- surd. Therefore the point of intersection of E B and D F is not in the circumference of the circle K, and therefore, a line, H C, drawn from this point to the centre C, does not cut off a third part B I of the arc A B ; because If B 1 be a third part of the arc A B, the locus of the intersection of C I with E B is the circumference of a circle passing through B, whose radius B K is equal to B C. Let B C I be the one-third of B C A, and let C I and E B produc- ed meet any where in H ; (it is easy to prove, that they will meet in that direction :) join E I, and through B draw B K || to E I. /CKB = CIE = CEI = |ACI=ICB.-.BC=BK. Again ^CKB = KCB = 2lEB = 2KBH;but ZCKB=KBH-f.KHB.-. KBH = KHB.-. KH=KB, and .-. K is the centre of a circle passing through H and B, and having its radius B K = to B C. The Major, therefore, has not trisected the angle : I believe, the problem remains just as it was before the Major took it in hand. I hope I may be allowed to correct an error, relating to the present subject, into wh-ch many fall, who have not leisure to examine geome- trical theories, &c, but yet have occasion to employ them. To tri- sect an angle, they are directed by some treatises, (Adam's Geometrical Essays, I believe, is onej) to find an arc that is equal to the third part of the given arc, and to lay it off on the given arc ; and a cumbersome mode is taught of finding this required arc, which is easily obtained thus : Let D A F be the given angle ; take AD = any length = 3 (for instance) ; and take A B = | AD=1, and describe the circles D F E, B G C ; then according to Newton's 5th lemma, A B~: A D : : arc B G : arc B F .-. arc B G = §■ arc D F. Now the diffi- culty is to lay off the arc B G upon the arc D F, so as to divide D F in- to three equal arcs. Recourse is had to a pair of compasses ; but it is very plain, that the compasses measure only the chord of B G, which is a straight line, and not the arc itself, which is a curve. Let the right angle D A E be trisected after this manner, and it will be found, that the chord B C will not answer the purpose. If it do, each por- tion of the arc being equal to 30% BC = V~2 =1.41, &c. will be the chord of 30° ; whereas the chord of 30° to the radius 3 is equal to 1.55, &c.