Circles with common tangent I am trying to solve this simple geometry problem but I am always tangled in so many equations it makes my head spin.  I tried solving it via similar triangles but i cant seem to eliminate all the unwanted variables. Please help.
I have to prove $ r_1\times r_3=(r_2)^2$

Thank you
 A: Due to the way they are constructed, the triangles $ABO,ACP,ADQ$ are similar, the figures $ABCPO$, $ACDQP$ are homothetic and $\dfrac{r_2}{r_1}=\dfrac{r_3}{r_2}.$
A: 
We have
\begin{align} 
\triangle AOB:\quad
|AO| &= \frac{r_1}{\sin\phi}
,\\
|AP|&=|AO|+|OP|=\frac{r_1}{\sin\phi}+r_1+r_2
,\\
\triangle APC:\quad
|AP| &= \frac{r_2}{\sin\phi}
,
\end{align}
which gives 
\begin{align} 
r_2&=
\frac{r_1(1+\sin\phi)}{1-\sin\phi}
.
\end{align}  
Similarly,
\begin{align} 
|AQ| &= |AP|+|PQ|=\frac{r_1}{\sin\phi}+r_1+2r_2+r_3
\\
&=
\frac{r_1}{\sin\phi}+r_1+
\frac{2r_1(1+\sin\phi)}{1-\sin\phi}
+r_3
,\\
\triangle AQD:\quad
|AQ| &= \frac{r_3}{\sin\phi}
,\\
\end{align}
hence
\begin{align} 
r_3&=
\frac{r_1(1+\sin\phi)^2}{(1-\sin\phi)^2}
,\\
\end{align}
and 
\begin{align} 
r_1r_3&=r_2^2
\end{align}
follows.
A: 
We have $h'$s of contacting circles from common apex A, the series of similar triangles:
$$  h_2-h_1=2r_1,  h_3-h_2=2r_2,  h_4-h_3=2r_3 \tag1$$
$$ \sin \alpha = \frac {(h_2-h_1)/2}{(h_2+h_1)/2}= \frac {h_2-h_1}{h_2+h_1} =\frac {h_3-h_2}{h_3+h_2}\tag2$$
Apply componendo/dividendo to each pair
$$\frac{1+\sin \alpha}{1-\sin \alpha}= \frac{h_2}{h_1} =  \frac{h_3}{h_2} =\frac{h_4}{h_3} = = k \tag3$$
which gives successive geometric means that are also called powers of circle corresponding to extreme $h$ values product of each pair:
$$ r_1=\sqrt{h_1h_2},\, r_2=\sqrt{h_2h_3},\, r_3=\sqrt{h_3h_4}  \, \tag4$$
$$ \frac{r_2^2 }{r_1 r_3}=\frac{h_2 h_3}{\sqrt{h_1h_2h_3h_4}}=\sqrt{\frac{h_2 h_3}{h_1 h_4}}=\sqrt{k/k}=1 \tag5$$
