Question

The bisector of angle B of an isosceles triangle ABC with AB=AC meets the circumcircle of triangle(ABC) at P. If AP and BC produced meet at Q, prove that CQ=CA

Answer 1

construction join pc
∠cbp=(1/2)∠abc
⇒2∠cbp=∠abc....1
∠cbp=∠cap....2(angles inscribed on the same base)
from 1 n 2 
∠bca=∠caq+∠cqa
∴∠cqa=∠bca-∠caq
⇒∠abc-∠cap    
2∠cap-∠cap   
=  ∠cap     
∠cqa=∠cap
∴cq=ca