to Five Points in a Plane. 



461 



ABC, and two points O, 0'. If in the triangle ABC, by- 

 means of the point 0, we inscribe a triangle A'B'C', and in the 

 triangle A'B'C, by means of the point 0', we inscribe a triangle 

 a/^7, then the triangles ABC, ccfiy are in perspective, viz. the 

 lines A a, B/3, C7 will meet in a point. 



This is very easily proved analytically; in fact, taking x = 0, 

 y-=0, z=0 for the equations of the lines B'C, C'A', A'B ; respec- 

 tively, and (X, Y, Z) for the coordinates of the point 0, then the 

 coordinates of (A, B, C) are found to be (-X, Y, Z), (X, - Y, Z), 

 (X, Y, — Z) respectively. Moreover, if (X', Y', Z') are the co- 

 ordinates of the point 0', then the coordinates of (a, /3, 7) are 

 found to be 



(0, Y', ZO, (X', 0, Y), (V, V, 0) 



Hence the equations of the lines Aa, B/3, C7 are 



x , y , z 



respectively, 

 respectively 



x , y } z = 0, x , y, z = 0, 



-X, Y, Z X, -Y, Z X,Y, 



0, V, Z' X', 0, Z' X', Y', 



that is, 



x{ YZ'-Y'Z)+y( Z'X)+*(-XY' ) = 0, 



#(-YZ' )+y( ZX'-Z'X) + x?( X'Y)=0, 



x{ + Y'Z)+y(-ZX' )+z{ XY'-X'Y) = 0, 



which are obviously the equations of three lines which meet in a 

 point. 



But the theorem may be exhibited as a theorem relating to a 

 quadrangle 1234 and a point O f ; for writing 1, 2, 3, 4 in 

 place of A, B, C, 0, the triangle A'B'C is in fact the triangle 

 formed by the three centres 41.23, 42.31, 43.12 of the qua- 

 drangle 1234, hence the triangle in question must be similarly 

 related to each of the four triangles 423, 431, 412, 123; or, 

 forming the diagram 



=0; 



41.23 



P 



4 



Q 

 3 



R 

 2 



S 

 1 



42.31 



3 



4 



1 



2 



43.12 



2 



1 



4 



3, 



we have the following form of the theorem : viz. the lines 



a4, /33, 72 meet in a point P, 



«3, /34, 7 1 „ „ Q, 



«2, £1, 7 4 n „ R, 



al, /32, 7 3 „ „ S, 



