Four kissing circles How can one go about solving the following problem?

Inscribe a circle in an arbitrary triangle. Call it's radius $r_1$.
  Inscribe three more circles so that each one is tangent to two sides of
  the triangle and the first circle (i.e., each at a different corner). Call the
  radii $r_2, r_3, r_4$. Find a relationship between $r_1, r_2, r_3$ and
  $r_4$.

The most promising method of attack for me was to consider the isosceles triangles at each corner: the base being the tangent line to the point of intersection of the angle bisector of the triangle and the first circle. But I'm stuck.
Any suggestions much appreciated.

I'm using the same notation as in the link mentioned by David Mitra. By similarity we have
$$
\frac{r}{OA}=\frac{r_a}{OA-r-r_a}=\sin\left(\frac{\alpha}{2}\right) 
$$
where $\alpha=\angle A$. Then
$$
\frac{r_a}{r} = \frac{1-\sin(\alpha/2)}{1+\sin(\alpha/2)}
= \frac{1-\cos((\beta+\gamma)/2)}{1+\cos((\beta+\gamma)/2)}
= \tan\left(\frac{\beta+\gamma}{4}\right)^2
$$
where we used that $\alpha+\beta+\gamma=\pi$. Similarly
$$
\frac{r_b}{r} = \tan\left(\frac{\alpha+\gamma}{4}\right)^2
\qquad\qquad
\frac{r_c}{r} = \tan\left(\frac{\alpha+\beta}{4}\right)^2
$$
Therefore, $\sqrt{r_ar_b}+\sqrt{r_ar_c}+\sqrt{r_br_c}$ can be written as
$$
\begin{split}
\frac{\sqrt{r_ar_b}+\sqrt{r_ar_c}+\sqrt{r_br_c}}{r}
= \frac{\sin(\rho)\sin(\sigma)\cos(\tau)+\sin(\rho)\cos(\sigma)\sin(\tau)
+\cos(\rho)\sin(\sigma)\sin(\tau)}{\cos(\rho)\cos(\sigma)\cos(\tau)}
\end{split}
$$ 
where $\rho=(\beta+\gamma)/4$, $\sigma=(\alpha+\gamma)/4$, and $\tau=(\alpha+\beta)/4$. Note that $\rho+\sigma+\gamma=\pi/2$.
Consider $x$, $y$, $z$ such that $x+y+z=\pi/2$. Then
$$
\begin{split}
\sin(x)\sin(y)\cos(z) &= \frac{1}{2}(\cos(x-y)-\cos(x+y))\cos(z) 
\\&=
\frac{1}{4}(\cos(x-y+z)+\cos(x-y-z) -\cos(x+y+z)-\cos(x+y-z))
\\&= \frac{1}{4}( \cos\left(\frac{\pi}{2}-2y\right) + \cos\left(-\frac{\pi}{2}+2x\right) - \cos\left(\frac{\pi}{2}\right)-\cos\left(\frac{\pi}{2}-2z\right)
\\&=\frac{1}{4}\left( \sin(2y)+\sin(2x)-\sin(2z)\right)
\end{split}
$$
And similarly
$$
\begin{split}
\cos(x)\cos(y)\cos(z) &= \frac{1}{2}(\cos(x-y)+\cos(x+y))\cos(z) 
\\&=
\frac{1}{4}(\cos(x-y+z)+\cos(x-y-z) +\cos(x+y+z)+\cos(x+y-z))
\\&=\frac{1}{4}\left( \sin(2y)+\sin(2x)+\sin(2z)\right)
\end{split}
$$
This means that $(\sqrt{r_ar_b}+\sqrt{r_ar_c}+\sqrt{r_br_c})/r$ is equal to
$$
\frac{\sqrt{r_ar_b}+\sqrt{r_ar_c}+\sqrt{r_br_c}}{r} = \frac{\sin(2\rho)+\sin(2\sigma)+\sin(2\tau)}{\sin(2\rho)+\sin(2\sigma)+\sin(2\tau)}=1
$$
