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

I’d like to suggest the following as a solution to a special case of this problem.
This special case requires an additional condition :- (C, A, D are on the same straight line.)
Note: Lines and markings of angles are shown in the figures attached.

Fact-1: $\Delta PBD \sim \Delta PDY$ [See figure 2.]
Thus, requiring to prove $\Delta PCA \sim \Delta PBD$ is equivalent to prove $\Delta PCA \sim \Delta PDY$.
Fact-2: PCDB is a cyclic quadrilateral.
Proof: $(<CPD) + [<CBD] = (π – α – β) + [ α + β] = π$
Fact-3: [See figure 1] Let x be the length of a chord on a circle with radius R. If that chord subtends an angle θ at the circumference, then x = 2R sin θ.
Therefore, $CA = 2r_1 sin α$ and $YD = 2r_2 sin α$
In $\Delta PCA$ and $\Delta PDY$,
$Angle PDY = angle PBD = angle PCA$
$CA : YD = r_1 : r_2$
$PC : PD = r_1 : r_2$
Therefore, $\Delta PCA \sim \Delta PDY$
