If AC and BC are two equal chords, BA is produced to P and CP cuts the circle at T the how is CT:CB=CA:CP?

Question

I've been solving the following question,

If AC and BC are two equal chords of a circle. BA is produced to any point P and CP, when joined cuts the circle at T then show that $\dfrac{CT}{CB}=\dfrac{CA}{CP}$?

I made a rough diagram and found that $\triangle PTB \sim \triangle PAC$. But this only gives $\dfrac{PT}{PA}=\dfrac{PB}{PC}$. Any help will be appreciated.

Answer 1

Since $\triangle{PTB}$ and $\triangle{PAC}$ are similar, we have $$\frac{CA}{CP}=\frac{BT}{BP}\tag 1$$ By the way, we have $\angle{PBC}=\angle{BTC}$. (this is because $\angle{PBC}=\angle{BAC}=\angle{BTC}$.) Also, we have $\angle{PCB}=\angle{BCT}$. So, it follows that $\triangle{PBC}$ and $\triangle{BTC}$ are similar. So, we have $$\frac{CB}{BP}=\frac{CT}{BT},$$ i.e. $$\frac{BT}{BP}=\frac{CT}{CB}\tag 2$$ From $(1)(2)$, we have $$\frac{CA}{CP}=\frac{CT}{CB}$$ as desired.