552 Proceedings of the Royal Irkh Academp. 



Cor. 3. — When the hexagon a/3'ya'/3y' reduces to a point, the 

 circle of Cor. 2 will be called, from analogy, Lemoiae's first circle 

 of the hexagon ; its radius squared is equal to 



H^ + R""- 



Peoposition' XI. — The perpendiculars from the symmedian point of 

 a harmonic hexagon, on the sides of the hexagon, are proportional to the 

 sides. 



Bern. — The perpendiculars from K on the lines C'A, AB' (fig., 

 Prop. Tni.), are in the ratio of sin C'AK: sin KAB'; that is, as 

 sin C'AA' : smA'AB' or : : C'A' : A'B'; but C'A'-.A'B': : C'A : AB', 

 since the points C, A', B', A' form a haiTaonic system. Hence the 

 perpendiculars from ^ on the lines C'A, AB' are proportional to 

 these lines, &c. Hence the proposition is proved. 



Definition I. — If be the circumcentre of the hexagon, the 

 circle on OK as diameter is called the Brocard circle of the hexagon. 



Definition II. — If the sides of the harmonic hexagon be denoted 

 by a, h, c, d, e, f and the perpendiculars from K on three sides by 

 X, y, %, u, V, w; then the auxiliary angle ® determined by any of the 

 equations x = ^ a tan ©, y = -J- b tan 0, &c., is called the Brocard 

 angle of the hexagon. -^ 



Cor. — The radius of the cosine circle of the hexagon is 'equal 

 to R tan ©. 



Por, let fall the perpendicular KX on A' C (see fig., Prop, x.); 

 then 



KX ^ KU= sin ?7= sin A' A C=A'C^2R. 

 Hence 



KU^ 2R = KX ^ ^'C = i tan ; 

 therefore 



KU= R tan ®. 



Peoposition XII. — The perpendiculars from the centre of the eircum- 

 circle on the sides of a harmonic hexagon meet its Brocard circle in six 

 points, which connect concurrently in two loayswith the vertices of the 

 hexagon. 



Bern. — Let the points (Pig. 4) of intersection of the perpendicular 

 with the Brocard circle be L, M, N, P, Q, R. Join KZ, KM; then 

 because the angle KLO is right, KL is parallel to BA'. Hence ZZ' is 

 equal to the perpendicular from Kon BA. Similarly, ITM' is equal to 

 the perpendicular from K on A' C. Hence (Prop, xi.) ZZ' : MM' : : BZ' 

 : A'M'. Hence the triangles BZ'Z, A' MM are equiangular; there- 

 fore the angle BZZ' is equal to A' MM'. Hence, if the lines BZ and 

 A'M intersect in Vl, the four points 0, M, Z, Q, are concyclic ; there- 



