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

Here's a partial solution.

I'll revise notation a little bit and use coordinates to help set the stage. My circles have centers $H(-h,0)$ and $K(k,0)$ and respective radii $r$ and $s$. (Without loss of generality, we assume $r > s$; the case $r=s$ is left to the reader.) The circles meet at points $A(0,a)$ and $B(0,-a)$. The origin is $O$.
Drawing right triangles $\triangle HOA$ and $\triangle KOA$ (with hypotenuses of length $r$ and $s$), we can define angles $\theta$ and $\phi$ (at $H$ and $K$) such that
$$r \cos\theta = h \qquad r\sin\theta = a = s\sin\phi \qquad s \cos\phi = k$$

First we observe that, because $|PC|^2 + r^2 = |PH|^2$ and $|PD|^2 + s^2 = |PK|^2$, the proportionality condition on $|PC|$ and $|PD|$ implies an identical condition on $|PH|$ and $|PK|$.
$$\frac{|PC|}{|PD|} = \frac{r}{s} \qquad \implies \qquad \frac{|PH|}{|PK|} = \frac{r}{s} \qquad (1)$$
Consequently, $P$ lies on the circle defined by 
$$\frac{( x + h )^2 + y^2}{r^2} = \frac{(x-k)^2 + y^2}{s^2}$$ 
Let $Q(q,0)$ be the circle's center, and $p$ its radius; we can determine that 
$$q = \frac{h s^2 + k r^2}{r^2-s^2} = \frac{r s \; \left( r \cos\phi + s \cos\theta\right)}{r^2 - s^2} \qquad
p = \frac{r s\;\left(h+k\right)}{r^2-s^2} = \frac{rs\;\left( r \cos\theta + s \cos\phi\right)}{r^2 - s^2}$$
In right triangle $\triangle QOA$, we define angle $\psi$ at $Q$ such that
$$p \cos\psi = q \qquad p \sin\psi = a$$

Fact. $\psi = \phi - \theta$.
Proof. Show that $\cos\psi = \cos\left(\phi-\theta\right)$, expressing the trig quantities in terms of $h$, $k$, $r$, $s$, $a$, and invoking the relation $r^2 - h^2 = a^2 = s^2 - k^2$.

Now, since $\theta$, $\phi$, $\psi$ measure half of a central angle subtending chord $AB$ in respective circles $\bigcirc H$, $\bigcirc K$, $\bigcirc Q$, they also measure any inscribed angle in those circles subtending (and on the appropriate side of) the same chord. In particular,
$$\angle AEB = \theta \qquad \angle AFB = \phi \qquad \angle APB = \psi$$
where $E$ and $F$ are the "other" points where $\overleftrightarrow{PA}$ and $\overleftrightarrow{PB}$ meet $\bigcirc H$ and $\bigcirc K$.

Note that $\angle AFB = \phi$ is an exterior angle of $\triangle APF$, which has remote interior angle $\psi = \phi-\theta$; consequently, $\angle PAF = \theta$. This implies that $AF$ and $EB$ are parallel, so that $\triangle APF \sim \triangle EPB$. Thus, for instance,
$$\frac{|PA|}{|PF|}=\frac{|PE|}{|PB|} \qquad (2)$$
By the Power of a Point "secant-tangent" theorem applied to $P$ relative to $\bigcirc H$ and $\bigcirc K$, we have
$$|PA||PE| = |PC|^2 \qquad |PF||PB| = |PD|^2 \qquad (3)$$
Using $(3)$ to eliminate $|PE|$ and $|PF|$ from $(2)$ gives
$$\frac{|PA|}{|PD|^2/|PB|} = \frac{|PC|^2/|PA|}{|PB|}$$
whence
$$\frac{|PA|}{|PD|} = \frac{|PC|}{|PB|}$$
which completes two-thirds of the desired similarity proof. The final third requires demonstrating either that $|AC|/|BD|$ is equal to the above, or that $\angle APC \cong \angle BPD$. Arduous coordinate calculations seem to bear this out, but the process isn't at all pretty, so I'll leave things here (for now).
