Dr STEW ARTs 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(EAM-EBM-EC*-|-ED*) = 3(LA*-f-LBM-LO+U>)+ 

 3.4.EL 2 , (Theor. 6.) ; and, for the fame reafon, 

 4 (EX 2 +EY 2 +EZ*) - 4 (LX 2 +LY 2 +LZ i )+4. 3 .EL\ 

 But by conftruclion, 



3 (LA 1 +LB i +LC 2 +LD 2 ) rr 4(LX*+LYM-LZ 2 ). Therefore, 

 3 (EA 2 +EB 2 -f-EC 2 -fED 2 ) rz 4 (EX 2 +EY 2 +EZ 2 ). 



Cafe 2. When the lines (Fig. 13.) AB, AC, AD, AE, given 

 by pofition, interfect 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, 

 (Theor. 6.), and let AH, AK, be two lines, whofe pofition is 

 given, fuch, that 2(FB 2 +FC 2 +FD 2 +FE 2 ) c= 4 (FH 2 -fFK 2 ), 

 (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 bifefted in L. Join AX, AY, AZ, and FX, FY, FZ. 

 3(FB i +FC 2 +FD 2 +FE I ) = 6(FH'+FK 2 ), (Theor. 12.), and 

 4 (FX 2 +FY 2 +FZ 2 ) = 6(FH 2 -fFK 2 ) = 3(FB 2 +FC 2 -j-FD 2 

 -f-FE 2 ). Therefore AX, AY, AZ, are the three lines required 

 to be found. 



Cafe 3. When the lines (Fig. 14. No. 1.) AB, BC, CD, 

 DA, are not parallel, and do not interfect one another in the 

 fame point. 



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

 that it {hall be the centre of gravity of the four points L, M, 

 N, O, where they are interfecled by the perpendiculars XL, 

 XM, XN, XO, drawn to them from X, (Theor. 13.) j and 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, in 



P, 



