92 



Fig. I. 



DISCUSSION. 



(1) With P and Q as centers, and PD and QK as radii draw two 

 arcs meeting in B. Then PBQ is a right angle, and PB is tangent to 

 arc EBK, and as the tangents PD and PB to circle O and arc EBK are 

 equal, P must be on the common chord KE produced; and in the same 

 way DCQ is a straight line. 



(2) Since PK.PE=PF.PG=PQ.PA. the point A is in the circumference 

 T, and OA is perpendicular to PQ. and A is also in the circumference S. 



(3) PQ.PA=PF.PG=PI.PH. .-. the points A, Q, I, H are concyclic, 

 and in the same way A, P, I, F are also concyclic. 



(4) PK and QD are respectively perpendicular to QO and PO. and 

 R is the orthocenter of the triangle POQ, and AO passes through R. 



