' Dr STEW ART'S THEOREMS. 127 



It is plain that L is the centre of gravity of the points 

 X, Y, Z, and becaufe it is alfo the centre of gravity of the 



points A, B, C, D, 

 3(EA'+EB^+EC^+ED0 = 3(LA^+LB^+LC^+LD0+ 

 3 4 EL* (Theor. 6;) ; and, for the fame reafon, 

 4(EX'+EY^+EZ0 - 4(LX'+LY^+LZ0+4.3-EL'- 

 But by conftru^ion, 



.(LA^+LB^+LC^+LDO = 4(LX'+LY^+LZ0. Therefore, 

 3(EA^+EB^+EC'+ED0 =z 4(EX'+EY»+EZ = > 



Cafe 2. When the lines (Fig. .3-) AB, AC, AD, AE, given 

 by pofition, interfea one another in the fame point A. 



Let G be the centre of gravity of the four points B, C, D, E, 



in the circumference of the circle of which AF is the diameter, 



rTheor. 6.), and let AH, AK, be two lines, whofe pofation is 



given, fuch, that .(FB^+FC^+FD = +FE*) = 4(FH'+FK^), 



(Theor 12.)- From any point X in the circumference draw, 



through G, the line XGL, fo that XG = 2GL ; and through 



L draw YLZ to meet the circumference in Y, Z, and fo as to 



be bifeded in L. Join AX, AY, AZ, and FX, FY, FZ. 



^fFB'+FC^+FD'+FE') =6CFH'+FK0, (Theor. 12.), and 



1(FX^+FY=+FZ0 = 6(FH^+FK0 = 3(FB = +FC^-f-FD = 



+FEO. Therefore AX, AY, AZ, are the three lines required 



to be found. 



Cafe 3. When the lines (Fig. 14- No. i.) AB, BC, CD, 

 DA, are not parallel, and do not interfed one another m the 



fame point. 



Let X be a point fo related to the lines AB, BC, CD, DA, 

 that it Ihall be the centre of gravity of the four pomts L, M, 

 N O where they are interfeded by the perpendiculars XL, 

 XM, XN, XO, drawn to them from X, (Theor. I3-) J ^nd let 

 XP XQ, XR XS, be drawn from X parallel to AB, BC, CD, DA, 

 and let them meet the perpendiculars to thefe lines, from E, m 



