94 



P. D utt— - Properties of the circle and analogous matters. [No. 2, 



then ■=-= = -zrr^ 



The properly follows also from the consideration of the equality 

 of the areas of the triangles APO and BPO. 



Let AP cut the circle again in P', then Fig r 4. 



AP_ AT 



BP " BP' 



AP _ BP 



'*• AP'~ BP' 



AP BP BP . 



£P. 



.-. PQ is || AB. (7). 



Let PB produced meet the circle in R 



AP_ AR 



PB" BB 



AP_ BP 



'*' AB" BB 



.'. /.PAB^^BAR, (8). 

 Similarly it can be shewn that if P'B meet the circle again in R' 

 then 



£P'AB=ABAR'. 

 .'. A, R', B lie on the same straight line. (9). 

 Again, in the two triangles APB, AR'B we have Z.PAB- /.R'AB 

 also Z_APB = *- £1?'PR^tc-£P'R'R=£AR'B, and the side AB is 

 common 



.-. we have BP = BR', AP = AR'. 



Similarly we can show that BP' -BR and AP' ~ AB. (10). 

 Also BQ = BP. 



.: BP'*-BR'* = P'R' .P'Q. (11). 



Since BP = BQ = BR' we can show that the angle QPR' is a right- 

 angle. (12). 



If R'Q' be drawn parallel to AB, similnrly AQQ'R' is a right-angle, 

 and as PQ, R'Q' are each parallel to AB, QPR'Q' is a rectangle. (13). 

 And it is evident that P, Q, Q', &' h'e on a circle, the centre of 

 which is B. (14). 



The results of propositions (7) to (14) may be summed up as 

 follows : — 



If a straight line be drawn through the fixed point A, cutting the 

 circle in P and P' and PB, P'B meet the circle again in R and R', 

 and PQ and R'Q' are drawn parallel to AB, then A, R', R lie on one 

 straight line. A circle with centre B passes through P, Q, Q', R'. AB 

 bisects the angle P' AR' and P, R' and P', R are respectively symme- 

 trical with respect to AB. 



