The bisector of `/_B` of an isosceles triangle `A B C` with `A B=A C` meets the circumcircle of ` A B C` at `P` as shown in Figure. If `A P\ a n d\ B C` produced meet at `Q ,`. prove that `C Q=C A`

Explanation

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 angle PBC = /_QAC + /_ AQC`

Now,
`/_ QAC = /_ PAC` and because APAC and `APBC` 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.