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BULLETIN OF THE 



A D iu R ; then B R = iv. L. Hence from data the triangle 

 B M R is known. 



It is well known that 2aBMR + CM. DR = rect. A 6 = 

 2aBMR + 2aCLM. (lY) 



Fig. 4. 



Take B E = B L, AF=:AM; then aBKE = aBKL, 

 A A K F =A A K M, and a F K E = a L K M, because the angles 

 L K M, F K E, are supplementary ; therefore □ A M L B = 

 2 A A K B ; hence by (IT) A A K B = i A B M R. 



Construction. — Make a triangle B M R, having its sides B M, 

 B R, equal to the given bisectors, and the angle M B R equal to 

 half a right angle. On M R draw a semicircle, and construct a 

 hyperbola having B M, B R, for asymptotes, and such that the 

 rectangle under the ordinate and abscissa (parallel to the asymp- 

 totes) is half the rectangle under the given bisectors. Let this 

 hyperbola cut the semicircle in A ; join A B and produce A K 

 parallel to B R, so that A L == B R ; and produce B L, A M, to 

 meet in C. Then ABC will be the triangle required. 



BIBLIOGRAPHICAL NOTES AND ACKNOWLEDGMENTS. 



This problem was proposed in the Ladies' Diary for 1797, by 

 Alex. Rowe, and the following year two solutions of it were 

 given; one by William Burdon and the other by J. Hartley. 

 Our sixth solution is taken from Mr, Burdon, as published in 

 Leybourn (Thomas). The Mathematical Questions proposed in 

 the Ladies' Diary, etc., 8vo., London, 1817, vol. iii. 328. 



Mr. Hartley's solution is trigonometrical, the unknown quan- 

 tity being tan ^ A, and his final equation corresponds to equation 



