How prove this geometry $\Delta PCA \sim\Delta PBD$ let the circle $O_{1} $ and the circle $O_{2}$ the  radius of is $r_{1},r_{2}$ respectively,and the circle $O_{1}$and $O_{2}$ intersection with $A$ and $B$,and  the tangent to $O_{1}$ at $C$,and the tangent to $O_{2}$ at $D$, and such 
$\dfrac{PC}{PD}=\dfrac{r_{1}}{r_{2}}$, show that：
$\Delta PCA \sim\Delta PBD$

and This problem is my student ask me, and I find this same problem with 2012 china girls problem,But for my student problem, I can't prove it,and I guess this problem have some nice methods, Thank you everyone.
see this same probelm：http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2769659&sid=f359cf6f214ae30fda2709d047be0587#p2769659

Referring to the latter part of @Blue’s partial solution, I've found the following slightly simplified version.
From the figure provided by Blue, I add the following constructions:


*

*Produce EB to cut the x-axis at R.

*Join AR and join HB.
See  
Then, it is not so difficult to see that (i) AHBR is rhombus of length = r; and (ii) the four isolated triangles are similar.
In particular, $\Delta FAP \sim \Delta KRA$.
Thus, $\frac {PA} {PF} = \frac {AR} {AK} = \frac {r} {s}$
Then, we are still have to prove "The final third requires demonstrating either that |AC|/|BD| is equal to the above, or that ∠APC≅∠BPD."
Probably, I think the question will be solved if we can prove that β – α = ω.
