1832.] On the Trisection of Angles. 501 



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 



