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

Answer 2

Hello Mate!

Since AB = AC, hence angle opposite to equal sides are also equal.

< ACB = < ABC.

Since, ext < ACB = < QAC + < AQC

Then, < ABC = < QAC + < AQC

Now, because < ABP = < PBC.

Therefore, 2 < PBC = < ABC

Hence, 2 < PBC = < QAC + < AQC

Now, < QAC = < PAC and because ∆PAC and ∆PBC lie on same line segment PC.

So, < PAC = < PBC

Hence < PBC = < QAC

2 < PBC = < PBC + AQC

< PBC = < AQC

< PAC = < AQC

< QAC = < AQC

Side opposite to equal sides are equal.

QC = AC

Hence prove.

Have great future ahead!

Wishing you and your family a Happy New Year!