132 DEMONSTRATIONS of 



Let m be =: 3. Let D be the centre of gravity of the three 

 points A, B, C J join AD, BD, CD ; from E draw EF, EG, EH 

 perpendicular to AD, BD, CD; in AD take DK. = ^ AD, in 

 BD take DL = ^BD, and \a DC take DM = 7DC. Then» 

 AE' = DE^+AD^— 2AD.DF 

 BE' = DE'+BD^+2BD.DG 

 CE' = DE^+CD'— 2CD.DH. Therefore, 

 AE* = DE*+2AD^DE'— 4AD^DF+AD'»— 4DE^AD.DF-h 

 4AD\DF^ 



BE^ = DE'»+2BD^DE'+4BD^DG^-BD*+4DE^BD.DG+ 

 4BD\DG\ 



CE* rr DE*+2CD-.DE'— 4CD'.DH+CD*— 4DE.CD.DH+ 

 4CD'.DH\ But becaufe D is the centre of gravity of the three 

 points A, B, C, AD.DF+CD.DH = BD.DG. Therefore,, 

 making AE'^+BE^+CE* — S*, we ihall have 



fAD^DP] f— AD'.DF] fAD\DF'] fAD*} 

 S^=3DE*+2]rD\De4+4 +BD^DGf+4{BD^DG4+<BD'»L 

 ICD^DE'J I— CD'.DH) ICD\DH'J lCD*J 



But DE' = EF^+DF' = EG'+DG' = EH'+DH\ There- 

 fore, 



fAD^EFM fAD'DFM f— AD'.DF] [ADM 

 S'=:3DE♦+2]BD^EG4^-6 BD^DG4^-4j+BD^DG[+ BD* L 



lCD\EffJ lCD\DH'J I— CD^DhJ ICD"! 



rAD\EF' ] fAD'(6DF^— 4AD.DF+AD')] 

 Or, S* = 3DE*+2-'.BD\Eg4+ BD^(6DG'+4BD.DG+BD^)[. 

 IcD\Eh4 [cD'(6DH'— 4CD.DH+CD')i 



But 3DK = AD ; 3DL = BD, and 3DM = CD ; and confe- 

 quently 9DK' = AD% qDL* = BD', and gDM^ = CD\ 

 Therefore, S* = ^DE* + 



"fAD\EFn fAD»f6DF=— i2DK.DF-f-6DK0 1 r3DK\ADM 

 2<^BD\EG4+iHD {6DG'+i2DL.DG+6Di. ) +<3DL\BD= I. 

 lCD\EH»J lCD\6DH'— I2DM.DH+6DM')) l3DM\CDV 



