132 DEMONSTRATIONS of 



Let m be — 3. Let D be the centre of gravity of the three 

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

 perpendicular to AD, BD, CD; in AD take DK — \ \D, in 

 BD take DL = iBD, and ij DC take DM = fDC. Then, 

 AE 2 = DE 2 +AD 2 — 2AD.DF 

 BE 2 = DE 2 +BDH-2BD.DG 

 CE 2 — DE 2 -f-CD 2 — 2CD.DH. Therefore, 

 AE 4 = DE 4 -f-2AD\DE 2 — 4AD 3 .DF-r-AD 4 — 4 DE 2 .AD.DF-f- 

 4AD 2 .DF 2 . 



BE 4 = DE 4 +2BD 2 .DE 2 +4BD 3 .DG+BD 4 + 4 DE 2 .BD.DG+ 

 4BD\DG\ 



CE 4 = DE 4 +2CD 2 .DE 2 — 4CD'.DH+CD 4 — 4DE.CD.DH-f 

 4CD 2 .DH 2 . But becaufe D is the centre of gravity of the three 

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

 making AE 4 -f-BE 4 -fCE 4 = S 4 , we (hall have 



fAD 2 .DE 2 ] f— AD'.DF] [AD 2 .DF 2 ] fAD 4 

 S 4 =3DE 4 +2JRD 2 .DE 2 +4 +BD 3 .Dg[+ 4 ^BD 2 .DG 2 WbD 4 ' 

 lCD 2 .DE\) I— CD'.DHl lCD\DH*J I CD* 



But DE 2 = EF 2 -f DF 2 = EG 2 +DG 2 = EH 2 -f DH 2 . There- 

 fore, 



fAD 2 .EF 2 ) fAD 2 DF 2 1 f— AD 3 .DFj (AD 4 

 S 4 =3DE 4 +2]BD 2 .EG 2 +6 BD\DG4+ 4 ]+BD 3 .DGf+ BD 4 



ICD 2 .EH 2 J ICD\DH 2 J I— CD 3 .DHJ ICD 4 



AD 2 .EF 2 1 (AD 2 (6DF 2 — 4 AD.DF-f AD 2 )] 

 BD 2 .EG 2 }-f- BD*(6DG 2 -f 4BD.DG+BD 2 ) L 

 CD 2 .EH 2 J ICD 2 (6DH 2 — 4CD.DH-f-CD 2 )i 



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

 quently 9DK 2 = AD 2 , oDL* =: BD 2 , and 9DM 2 = CD 2 . 

 Therefore, S 4 = 3DE 4 -f 



Or, S 4 = 3DE 4 +2- 



2 



*AD 2 .EF 2 ") fAD 2 (6DF 2 - 

 BD 2 .EG 2 H- 

 .CD 2 .EH*J 



s — I2DK.DF-T-6DK') } f 3 DK\AD 2 ] 

 BD (6DG 2 -f-i2DL.DG-r-6DL) +< 3 DL 2 .BD* L 

 -CD 2 ^6DH 2 — 1 2DM.DH-I-6DM 2 ) I IjDiM'.CD'J: 



Or„ 



