Page:Journal of the Asiatic Society of Bengal Vol 1.djvu/555

lines of equal parts, which are cut upon the piece A D, and then the given angle will be trisected. Note. — Fig. 4 is a sketch of the instrument above alluded to : it is copied from the 1st volume of the Mechanic's Magazine. III. — On the Trisection of Angles. By Mr. W. Masters, Verulam Academy. In one of the late numbers of the United Service Journal, there is an article on the trisection of an angle, written by a Major in the British Army, resident at New South Wales. This gentleman be- lieves he has solved the problem, and demonstrated his solution, which is in substance as follows : Let A B C be the angle to be trisected ; from C as a centre, and with any radius C B describe a circle ; produce A C, B C, to E and D ; join E B ; trisect the arc A B in F ; join C F and D F, and pro- duce the line D F, till it meets the line E B produced in H ; join CH ; C H cuts the arc A B in I ; the arc B I is one-third of the arc A B : this is his solution of the problem. In the demonstration, he pro- ceeds to say, take B K=B C, and from the centre K, with the radius K B, describe a circle. &c. &c. The demonstration appears to be complete, and the gentleman exults at his success ; but it escapes his sagacity, that his demonstration is built upon the hypothesis, that the circle K passes through the point H. Now, this is a very important point ; the whole solution depends upon it : it is obtained by the intersection of the lines E B and D F produced. My object is to prove, that the locus of the point of intersection, of the two lines E B and D F, is not the circumference of a circle, which passes through B, and whose radius B K is equal to B C. If possible, let the point H be in the circumference of the circle K, described as stated above. • BK=KH = BC = CD. .-. BK+ KH = BC + CD = BD;but BK + KH>-BH.BD>.BH. Again z EBD = BHD + BDHand E B D = B C F, because E H || C F, .•.BCF^BHD + B D H ; but BCF = < V /BDF.-. BDH = BHD