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.
 A: 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
$$
\frac{\sqrt{r_ar_b}+\sqrt{r_ar_c}+\sqrt{r_br_c}}{r} = \frac{\sin(2\rho)+\sin(2\sigma)+\sin(2\tau)}{\sin(2\rho)+\sin(2\sigma)+\sin(2\tau)}=1
$$
