44. Let 2 be the Brocard Point at which AC subtends the supplement of A; and let a, B, y be such points taken in BC, CA, AB that the circumcircles of Aßy, Bya, Caß cut in 2*. Then (1) aßy is similar to BCA. (2) 2 is one of the Brocard Points of aßy. (3) The Brocard angle of aby is equal to that of BCA. (4) The angles Bla, CAB, ANy are equal (being the angles by which the figure BCA would have to rotate round 2 in order to take up the position of the similar figure aßy). 45. The cosine-centre is a Brocard Point of each of the triangles, similar to the primitive triangle, in which the Lemoine Circle cuts the sides :—the other Brocard Points of these triangles being the Brocard Points of the primitive triangle. 46. If X, Y are the feet of the perpendiculars from 0 on AB, AC; then XY is perpendicular to the antiparallel of AO with respect to BAC. 47. The centre of the circle which passes through the feet of the perpendiculars from the Brocard Points on the sides is the point of intersection of the line joining the Brocard Points with the line joining the circumcentre and cosine-centre. 48. The Brocard Circle passes through the middle points of the three chords of the circumcircle which join the cosine-centre to the vertices. * Thus: take a anywhere in BC, and let the circles Ba, NCa cut AB, AC respectively in y, B. CHAPTER X. TRIGONOMETRICAL FORMULÆ FOR CIRCLES AND RECTILINEAL FIGURES. a = § 1. CIRCLES AND TRIANGLES. 207. Expressions for the area (S) of a triangle. By Art. 117 the area of a triangle = half the product of two sides and the sine of the non-obtuse angle included by them. Now since the sine of an angle is equal to the sine of its supplement (Art. 137), the sine of the internal angle may be always used in this expression. Thus: given two sides (a, b) and the included angle (C), S= area of triangle = {ab sin C....... ...(1). c sin A csin B Now and b= sin C sin C .. substituting in (1), csin A sin B S = area of triangle = .(2). 2 sin C This form gives the area in terms of the angles and one side. Again, by Arts, 150, 152, (1) becomes S = area of triangle = {s (8 – a) (8-6) (8 – c)}...... (3). area of - BIC = {BC.IX = ža.r, area of A AIB = {AB. IZ = {c.r, Now AY = AZ; BZ = BX; CX=CY. :. AY+BX+CX = 1 (a + b + c) = 8. .. AY or AZ=8 - a; BZ or BX=8-6; CX or CY=8 – C. ::r=IY = AY tan IA Y = (8 – a) tan } A......... (2). Or again : BC sin IBC p=IX = IC sin ICX = sin ICX sin BIC (3). sin 1 (B+C) cos ? A 209. Expressions for the radius (r) of the escribed circle. Bisect the exterior angles at B and C by BI, CI Draw the perpendiculars 1,X1, I,Y, 1,2, on the sides. Then I,X,= 1,Y,=1,2,=r1. Then area of a BIC = 1 BC.1,X=fa.ri, area of AI,B=AB. 1,2, = }c.ru, .... (1). Now AY,= AZ1; BZ, =BX,; CX,=CY. .(2) .. AY, or AZ = 1 (a + b + c)= 8, and BX, or BZ =8-c; CX, or CY, = 8 – 6. ..^, = AY, tan I AY, = 8 tan 1 A..... = BX,cot BIX= (s - c) cot 1 B. Or again : BC sin BCI rn = 1,X, = BI, cos BI X, sin BIC cos BI X, cosA . :(3). 210. Expressions for the Radius (R) of the circumscribing circle. a a с Through B draw the diameter BA' of the circumscribing circle. Join A'C. Then the angle BCA' in a semicircle is a right-angle. And the angle BAʼC, which is in the same or the opposite segment to BAC is either equal or supplementary to A. ... sin BAC = sin A. BC .. sin A 2R 5 ..(1). sin .. d, which was used in Art. 145 for the value of each of these fractions, is the diameter of the circumscribing circle. 28 abc (2). bc' 48 The formulæ (1) may be also derived by using the figure of the circumscribing circle in the last chapter. COR. The distances of S from the sides of ABC are R cos A, Rcos B, R cos C. If A is obtuse, cos A is negative, and the arithmetic value of the distance of S from BC is – R cos A, S being on the side of BC remote from A. It will be found convenient to express distances in terms of R and ratios of the angles. = |