502 On the Trisection of Angles. [Nov. 



.-. 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. 



