Every triangulation of an $n$-sided polygon can be associated with a binary tree with $(n-2)$ nodes, since there are $(n-2)$ triangles in the triangulation. Fix a side of the polygon, then consider the triangle on that side. It can have one or two neighbours, left or right. The triangles with only one neighbour are the ones that share a side (or two, in the case of the starting triangle) with the perimeter of the polygon, and the triangles with no neighbours are the ones that share two sides with the perimeter of the polygon. On the other hand, every rooted binary tree with $(n-2)$ nodes (where a right child differs from a left child) gives a different triangulation.
Moreover, every rooted binary tree with $(n-2)$ nodes can be encoded as a string of $(2n-4)$ balanced parenthesis: assign the encoding "()" to every leaf, then recursively assign the encoding "(encoding-of-the-left-child) encoding-of-the-right-child" to every parent, until the root is reached. A couple of parenthesis is added every time an edge of the tree is traversed. Moreover, every string of $(2n-4)$ balanced parenthesis can be uniquely "decoded" into a rooted binary left/right tree with $(n-2)$ nodes. The last bijection is between the string of $(2n-4)$ balanced parenthesis and the up-or-right paths in a $(n-2)\times (n-2)$ square grid: starting from the lower left corner, we move up every time we encounter a "(" and to the right every time we encounter a ")": the balancing condition is equivalent to the fact that we never cross the diagonal of the grid. Now see the second proof on Wikipedia about Catalan numbers and everything is done.