Br STEWART'S THEOREMS. 115 



Since a. AH = J.BH, we mall have *.EK = £.FK ; and 

 fince («+J)HX ss c.CX, we (hall alfo have 0+£)KX = f.GX. 

 Therefore lince 



£.XF = £.FK— J.KX, and 



r.XG = (*+*} KX, we fhall have 



b.XV+c.XG = b.TK+a.KX = a.EK+a.KX = a.XE, and 



2DX (— a.XE-f-J.XF+r.XG) = o ; therefore 

 a.AD*+b.BD*+c.CD* - a.AX*+b.BX*+c.CX*+{a+b+c)T>X\ or 



AD*+ — BD*+— CD* = AX*+— BX'+ — CXM- (~^)bX\ 



1 a a 'a 'a * \ a / 



The point X is the centre of gravity of weights, proportional to 

 the magnitudes a, b, c, &c- placed at the given points A, B, C, 

 &c. 



Cor. 1. Let any number, m, of circles be given by pofition, 

 (fig. 3.) and about every circle let an equilateral figure be de- 

 fcribed, a point X may be found, fuch, that if from any point 

 C there be drawn perpendiculars to the fides of the figures, and 

 a flraight line to the point found, twice the fum of the fquares 

 of the perpendiculars will be equal to the multiple^of the fquare 

 of the line drawn to the point found, by the number of the 

 fides of the figures, together with a given fpace. 



Let m be rz 2 ; let a be the number of the fides of the figure de- 

 fcribed about the circle whofe centre is A, b the number of the 

 fides of the figure defcribed about the circle whofe centre is B, 

 CD, CE, CF, the perpendiculars to the fides of the firfl figure, 

 and CG, CH, CK, CL, the perpendiculars to the fides of the 

 fecond. 



Join the centres A, B, and divide AB in X, fo that AX : BX 

 — b: a, X will be the point required. 



2(CD*-f-CE*-f CF 2 ) == a«.AM a +a. AC 2 (theor. 3.). In like manner, 

 2(CGM-CH*+CK>+CL*) - 2 £.BNH-£.BC*. Therefore, 

 2(CD*+CE*+CF*-fCG*-f CH*-fCK 2 -f-CL 2 ) - 2 ^.AM 2 -f 2£.BN* 

 +*.AO+J.BO. But, 



p 2 a.AC 2 



