Mr. J. J. Sylvester on a simple Geometrical Problem. 367 



AD and BE intersect in H, it is readily demonstrable, by a 

 suitably constructed apparatus of similar triangles, tliat 

 AH:BH::CE:CD. 

 But as HBA is less than HAB, AH is less than BH, and 

 therefore CE is less than CD, and therefore CED is greater than 

 CDE ; that is to say, CAB less K is greater than CBA plus K, 

 and therefore DAB less K is greater than EBA, i. e. ADE is 

 greater than ABE, and therefore the perpendicular from A upon 

 DE is greater than that from E on AB, which is easily proved 

 to be absurd. Hence, as before, the triangle is proved to be 

 isosceles. This proof, it is obvious, remains good for all cases in 

 which EB and DA, drawn on either side of the base, divide the 

 angles at the base proportionally, provided that these lines re- 

 main equal, and make positive or negative angles with the base 

 not less than one- half of the respective corresponding angles 

 which the sides of the triangle are supposed to make with it. 

 The analytical solution of the question, as might be expected, 

 extends the result still further. To obtain this, let 

 B AC = n . BAD, ABC = n . ABE, 

 (n) for the present being any numerical quantity, positive or ne- 

 gative; calling BAC = 2n«, ABC = 2n/3, we readily obtain, by 

 comparison of the equal dividing lines with the base of the triangle, 



sin(2?ia + 2/6) _ sin(2nj5 + 2a ) 

 sin2«« ~~ sin2ra/3 ' 

 or sin (2»« + 2/3) _ sin 2na. _ 



sin(2?j/8 + 2«) ~ sin2H^' 



and by an obvious reduction, 



tan(n-l)(«-/3) ^ tan (» + l)(a + ;g) 

 tan?i(a— /8) ~ tann(a + /3) 

 When this equation is put under an integer form, it is of course 

 satisfied by making « = /3; on any other supposition than a = fi 

 it c\idently cannot be satisfied by admissible values of the angles 

 for any value of n between + 1 and + oo ; for on that supposi- 



180 

 tion, since {«■ — ft) and (« + /3) arc each less than ~^, the first 



side of the equation will be necessarily a proper fraction and 

 positive ; but tlie second side, either a positive improper fraction 

 if (n + l)(« + /3) be less, and a negative proper or a negative 

 improper fraction if («+ l)(a + /S) be greater than a right angle. 

 If n be negative, let it equal — v, then 



tan(v+l)(«-/3) _ tan(v-l)(«-f/g) , 

 tan>/(«-/a) tauv(«-|-/S) 



