First of all, I need to remark that the current formulation of the question doesn't seem to be quite right, since the fact that you're working on a circle isn't really used. It also isn't very clear what you mean by "joining into pairs.
However, if I read this question in the following way, it seems to work: We choose $2n$ arbitrary points on a circle. How many ways are there of drawing $n$ lines between $2$ points, such that there are no intersections between these lines inside the circle?
How to prove this? It is easiest to follow what's going on by drawing a few examples and keeping these nearby while reading this. We will denote the number of possibilities given $n$ as $Q_n$.
Now suppose we have $n$ fixed, as well as $2n$ points on a circle. We will fix one point $p_1$ and then give names clockwise: the first point one encounters when walking along the circle, starting from $p_1$, is called $p_2$, etc. The first thing now is to consider to which points $p_1$ may be connected. We remark that if we draw a line between $p_1$ and $p_i$, this divides the remaining points in the sets $X_1=\{p_2,\ldots,p_{i-1}\}$ and $X_2=\{p_{i+1},\ldots,p_{2n}\}$, each corresponding to the points on one side of the line. Since no two lines may intersect within the circle, this imposes a restriction on the possible points $p_1$ may be connected to. Namely, this needs be done in such a way that both sets $X_i$ have an even number of points. That is, we may connect the point $p_1$ to each point $p_{2j}$ with $1\leq j \leq n$. This can easily be seen in a drawing.
Now, what happens when we join up the point $p_1$ with one such point? We have a set $X_1$ of $2n_1$ points lying on a circle that need to be pairwise connected in such a way that no lines intersect, while the same holds for the $2n_2$ points in $X_2$. We therefore notice that if we connect the point $p_1$ to $p_i$, the remaining number of possible line configurations becomes $Q_{n_1}Q_{n_2}$.
The result can now be obtained by adding the numbers of possible configurations corresponding to each line from $p_1$ to some $p_i$. What we get, is that $Q_n = Q_{n-1} + Q_1Q_{n-2}+...+Q_{n-2}Q_1+Q_{n-1}$. It is best to first write it this particular way to avoid any problems `in the middle' when $n$ is odd. If we define $Q_0$ as $1$ (one could argue that there is exactly one way not to draw any lines inside a circle at all), we can express every $Q_n$ by the recurrence relation
$Q_0=1,\qquad Q_n=\sum_{j=0}^{n-1}Q_jQ_{(n-1)-j}.$
I don't know how Catalan numbers have been defined in your case, but this recurrence relation describes them. If you don't know this yet, there must be plenty of sources out there to prove it for you.
The two paths enclose an array of unit squares forming a convex polyomino. (Note that in this context convex has a special meaning.) Say that the region has $m$ columns, with $c_k$ squares in column $k$. For $k=1,\ldots,m-1$ let $r_k$ be the number of rows shared by columns $k$ and $k+1$. The upper righthand corner is at the point $\langle m,n+1-m\rangle$.
The idea is to use these numbers to construct a Dyck path (mountain range) of length $2n$. It will have $m$ peaks, of heights $c_1,\ldots,c_m$ from left to right. For $k=1,\ldots,m-1$ the descent after peak $k$ will have length $c_k-r_k+1$, while the descent after peak $m$ will of course have length $c_m$.
The valley between peak $k$ and peak $k+1$ will have height $c_k-(c_k-r_k+1)=r_k-1$, so the ascent from it to peak $k+1$ will have length $c_{k+1}-r_k+1$.
Now $c_{k+1}-r_k$ is the number of squares by which the top of column $k+1$ rises above the top of column $k$, so
$$c_1+\sum_{k=1}^{m-1}(c_{k+1}-r_k)+c_m$$
is the height above the baseline of the top of column $m$, which is $n+1-m$, and
$$c_1+\sum_{k=1}^{m-1}(c_{k+1}-r_k+1)+c_m=n+1-m+(m-1)=n\;,$$
as desired.
I leave it to you to check that the total descents is also $n$, that the path never drops below the baseline, and that the convex polyomino and hence the original two paths can be recovered from the Dyck path.
Best Answer
There is a nice recursive procedure for constructing the triangulation corresponding to a balanced parenthesis string.
Suppose that $\sigma$ is a balanced string of $n$ pairs of parentheses; it must have the form $(\sigma_0)\sigma_1$, where $\sigma_0$ and $\sigma_1$ are balanced strings, either of which may be empty, and the parentheses are a matched pair. Now start with an $(n+2)$-gon that has not yet been triangulated, and fix a pair of adjacent vertices $x$ and $y$, with $x$ clockwise from $y$. Label the remaining $n$ vertices $v_0$ through $v_{n-1}$ counting clockwise from $x$.
If $\sigma_0$ contains $k$ pairs of parentheses, draw lines from $x$ and $y$ to $v_k$ to create a triangle $xv_ky$. (If $\sigma_0$ is empty, $k=0$, and the line from $x$ to $v_0$ is already present. Similarly, if $\sigma_1$ is empty, then $k=n-1$, and the line from $y$ to $v_{n-1}$ is already present.) If $1\le k\le n-2$, we have partitioned the $(n+2)$-gon into a $(k+2)$-gon with vertices $x,v_0,\ldots,v_k$, the triangle $xv_ky$, and an $(n-k+1)$-gon with vertices $y,v_k,\ldots,v_{n-1}$. (If $k=0$ or $k=n-1$, one of the polygons is empty, and we have only the triangle and one polygon.)
Now repeat the process with the new, smaller polygon(s), taking the edges $xv_k$ and $yv_k$ as the baseline edges analogous to the edge $xy$ of the original polygon and using the parenthesis strings $\sigma_0$ and $\sigma_1$, respectively, to determine the vertex of the triangle erected on each of those baselines.