102 
(b) Second proof that P,, P,, and P, lie in a straight line. 
Draw PC and PA (Fig. 1). 
Now /s PP, Cand PP, Care right Zs. 
.*. P, P,, Cand P, are concyclic with PC as diameter. 
2 tARB= 7 PCB y PCP; 
and YO PE, Py 7 OPE, 
vit PAB—/ PAP ==PP. Ps 
Now P, P., A, P, are concyclic. 
See ig ee ee 
Sot Hee ib ee ea es 
.*. P, P, passes through P, and the three points are collinear. 
2. If PP, be produced until it intersects the circumcircle of (, ABC, at the 
point U,, then AU, is || to Simson’s line of P. (Fig. 14 
Now the points P, P,, P,, and © are concyclie. 
Snes ae ete Py Pate: 
But Z P, PC— Z U, AC, (are CU, common to both) 
and 7 PP, 6=U,-ACc; 
