Prove a geometry question about angles and radii in five collinear circles? Our teacher give us this question and I worked on it but I couldn't find a way to prove that. Is it possible to help me to prove that? 
Thanks.


Assumptions
I'm assuming that the following are true:


*

*$A$, $B$, $G$, $H$, $I$, $K$ are collinear

*$B$, $G$, $H$, $I$, $K$ are the centres of their respective circles

*$A$, $C$, $D$, $E$, $F$, $J$ are collinear and the line on which they lie is tangent to all the circles.


Definitions
We know these angles:


*

*$\angle CAB$ = $\angle DAG$ = $\angle EAH$ = $\angle FAI$ = $\angle JAK$. Call this angle $\theta$.

*$\angle ACB$ = $\angle ADG$ = $\angle AEH$ = $\angle AFI$ = $\angle AJK$ = $90^{\circ}$.


And we can define some lengths. Let:


*

*$R_1$, $R_2$, etc. be the lengths of $CB$, $DG$, etc.

*$H_1$, $H_2$, etc. be the lengths of $AB$, $AG$, etc.


Known Information


*

*For right angled trianges, we know that $\sin(\text{angle}) = \frac{\text{opposite}}{\text{hypoteneuse}}$, so:


$$\forall n \in \{1, \dots, 5\}, \sin\theta = \frac{R_n}{H_n} \Rightarrow H_n = \frac{R_n}{\sin\theta}$$


*

*Looking at the diagram, we can also see that:


$$\forall n \in \{1, \dots, 4 \}, H_{n+1} = H_n + (R_n + R_{n+1})$$
Forming a Rule
We can substitute for $H_n$ from the first equation into the second and rearrange:
\begin{align}
&& \frac{R_{n+1}}{\sin\theta} &= \frac{R_n}{\sin\theta} + R_n + R_{n+1} \\
\Rightarrow && R_{n+1} \left(\frac{1}{\sin\theta} - 1\right) &= R_n \left(\frac{1}{\sin\theta} + 1\right) \\
\Rightarrow && R_{n+1} \left(1 - \sin\theta\right) &= R_n \left(1 + \sin\theta \right) \\
\Rightarrow && R_{n+1} &= R_n \left(\frac{1 + \sin\theta}{1 - \sin\theta}\right) \\
\Rightarrow &&R_{n+k} &= R_n {\left(\frac{1 + \sin\theta}{1 - \sin\theta}\right)}^k \\
\end{align}
The result
This means that:
\begin{align}
R_1 \cdot R_5 &= R_1 \cdot \left(R_1 \cdot \left(\frac{1 + \sin\theta}{1 - \sin\theta}\right)^4\right) \\
&= \left(R_1 \cdot \left(\frac{1 + \sin\theta}{1 - \sin\theta}\right)^2\right) \cdot \left(R_1 \cdot \left(\frac{1 + \sin\theta}{1 - \sin\theta}\right)^2\right) \\
&= R_3 \cdot R_3
\end{align}
Extra results
This rule, coupled with the fact that there's nothing special about $R_1$ (it could just as easily have been called $R_0$ or $R_{1000}$) means that if we take any two collections of integers $A = \{A_1, A_2, \dots\}$ and $B = \{B_1, B_2, \dots\}$ and define $S_A = \sum_{a \in A}a$ and $S_B = \sum_{b \in B}b$ we get:
\begin{align}
\prod_{a \in A} R_{a} &= R_0^{|A|} \left(\frac{1 + \sin\theta}{1 - \sin\theta}\right)^{S_A} \\
&= R_0^{|A| - |B|} \left(\frac{1 + \sin\theta}{1 - \sin\theta}\right)^{S_A - S_B} R_0^{|B|} \left(\frac{1 + \sin\theta}{1 - \sin\theta}\right)^{S_B} \\
&= R_0^{|A| - |B|} \left(\frac{1 + \sin\theta}{1 - \sin\theta}\right)^{S_A - S_B} \prod_{b \in B} R_{b}
\end{align}
If $|A| \neq |B|$ and $z = \frac{S_A - S_B}{|A|-|B|}$ is an integer, this simplifies to:
$$
\prod_{a \in A} R_{a} = R_z^{|A| - |B|} \prod_{b \in B} R_{b}
$$
Or if $|A| = |B|$ and $S_A = S_B$ (i.e. two equal sized collections with equal sums), we end up with:
$$
\prod_{a \in A} R_{a} = \prod_{b \in B} R_{b}
$$
and your question is a special case of this where $A = \{1,5\}$ and $B = \{3,3\}$.
