Visualizing the parse structure of a range concatenation grammar
Edit: answer obeying the "spirit" of the question:

Let's use the rules
$$S(xy)\to A(x,y)$$
$$A(x,ayb)\to A(x,x)A(y,y)$$
$$A(x,x)\to\epsilon$$
Then we can derive $S(aabb)\to A(a,abb)\to A(a,a)A(b,b)\to\epsilon$.
In this case the parse tree is:


  $$
\begin{eqnarray*}
&amp;S(aabb)\\
&amp;\downarrow \\
&amp;A(a,abb)\\
&amp;\swarrow\quad\searrow \\
&amp;A(a,a)\quad A(b,b)\\
&amp;\searrow\quad\swarrow \\
&amp;\epsilon\\
\end{eqnarray*}
$$


Original answer obeying the "letter" of the question:

Using @MarzioDeBiasi's grammar
$$\begin{eqnarray*}S(xyz)&amp;{\color{red}\rightarrow}&amp;A(x,y,z),\\
A(ax,by,cz)&amp;{\color{green}\rightarrow}&amp;A(x,y,z),\\
A(a,b,c)&amp;{\color{blue}\rightarrow}&amp;\epsilon,\end{eqnarray*}$$
the picture is:


  $$\begin{eqnarray*}
&amp;S(aaaaabbbbbccccc)\\
&amp;{\color{red}\downarrow}\\
&amp;A(aaaaa,bbbbb,ccccc)\\
&amp;{\color{green}\downarrow}\\
&amp;A(aaaa,bbbb,cccc)\\
&amp;{\color{green}\downarrow}\\
&amp;A(aaa,bbb,ccc)\\
&amp;{\color{green}\downarrow}\\
&amp;A(aa,bb,cc)\\
&amp;{\color{green}\downarrow}\\
&amp;A(a,b,c)\\
&amp;{\color{blue}\downarrow}\\
&amp;\epsilon\end{eqnarray*}$$