120 DEMONSTRATIONS of 



the centre of gravity of the figure, will have a given ratio to the 

 femidiameter, and will make given angles with the femidiameterr y 

 drawn to the angular points of the figure. 



The centre of gravity of the figure ABCD is found by bt- 

 fecling AB in F, by joining FC and dividing it in G, fo that 

 CG = 2GF, and by joining GD and dividing it in H, fo that 

 DH — 3HG. Hence, and by joining BD and CA, the lemma 

 will be manifeft. 



For the triangle BFE is right-angled in F, and the angle BEF =z 

 ADB, is given. Therefore the ratio of BE, or CE, to EF is 

 given. 



Again, in the triangle CEF, the angle CEF = BEC-f BEF = 

 2BDC-f-ADB — a given angle ', and fince the ratio of CE to EF, 

 and of CG to GF are given, the line EG will divide the triangle 

 CFE into two triangles given in fpecies. Therefore the angle 

 CEG, and the ratio of CE, or DE, to EG, are given. 



Lastly, in the triangle DEG, the angle DEG= sDAC-f-CEG,, 

 is given ; and fince the ratio of DE to EG, and of DH to HG, 

 are given, the line EH will divide the triangle DEG into two* 

 triangles given in fpecies. Therefore the angle DEH, and the 

 ratio of DE to EH will be alfo given, 



THEOREM XII. Fig. VIII. 



Let there be any number, m, offraight lines AB, AG, AD, AE, &c; 

 given by pofition, interfiling one another in the point A, two Jlraight 

 lines AX, AY, may be found, which will be given by pofition, fuch y 

 that if from any point F there be drawn the perpendiculars FB, FC, 

 FD, FE, &c to AB, AC, AD, AE, &c and FX, FY, perpendicu- 

 lar to AX, AY, 



2(FB J -fFC 2 +FD 2 -r-FE 2 &c.) = w(FX 2 +FY 2 ). 



Let 



