Dr STEWART'S THEOREMS. 121 



Let m be = 4. Let G be the centre of the circle which 

 pafTes through A, B, C, D, E, F, and H the centre of gravity 

 of the figure BCDE. Join GH, and through H draw XHY 

 perpendicular to GH, meeting the circumference in X, Y, and 

 join GB, GC, GD, GE ; HB, HC, HD, HE, HF ; AX, AY, 

 FX, FY. Then, by Theor. 6. GB*+GO-f GD*-f GE* = 

 4GB 2 = HBM-HCH- HD*+HEH-4HG\ 

 But 4GB 2 - 4 GX* = 4(GH*-f XH<). Therefore alfo, 

 HB*+HC a +HD a +HE*+4HG* - 4 (HG 2 -f-HX 2 ); or, 

 HB*-f-HC 2 +HD -f-HE* = 4HX*. Again, by Theor. 6. 

 FB*+FC 2 +FD*+FE* - HB*+HCM-HD 2 +HE*-f 4FH*, and 

 therefore, 



FB a +FC 2 +FD 2 +FE 2 = 4(FH 2 -f-HX*). That is, 

 2(FBM-FCM-FD*+FE a ) = 8(FH a -fHX*) == 4(FX*+FY*), 

 (Prop. 1.). 



But becaufe the lines AB, AC, AD, AE, are given by pofi- 

 tion, the angles BAC, CAD, DAE, BAE, are given ; therefore 

 the angles BGC, CGD, DGE, BGE, which are the doubles of 

 them, are alfo given, and the ifofceles triangles BGC, CGD, 

 DGE, BGE, are given in fpecies. Confequently, the ratio of 

 the femidiameter GB to each of the lines BC, CD, DE, BE, is 

 given, and therefore the ratios of BC, CD, DE, BE, to one ano- 

 ther, are given ; and the angles of the figure BCDE are alfo 

 given, therefore the figure itfelf is given in fpecies. Therefore 

 (Lemma 3.) the ratio of GX to GH is given; and fince the 

 angle GHX is a right angle, the triangle GHX is given in fpe- 

 cies. Therefore the angles XGH, YGH, are given. But BGH 

 is given, (Lemma 3.) ; therefore BGX, BGY, and their halves 

 BAX, BAY, are alfo given; and fince BA is given by pofition, 

 and the point A, the lines AX, AY, are alfo given by pofi- 

 tion. 



Vol. II. q But 



