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$A_{1}A_{2}\dots A_{n}$ is a convex $n$-gon. How many ways are there to to divide the $n$-gon into $n-2$ triangles that have at least one segment in common with the $n$-gon using $n-3$ non-intersecting diagonals?

Because it is a bit hard I nearly can't find anything special, I want you to give hints and let me work by myself.

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    @bof Could you please give an answer now because I can't find the answer using hints.I find $\frac{n(n+3)}{2}.2^{n-5}$ but it isn't the answer in the book – 2017-01-11
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    Yes the book answer $n.2^{n-5}$ but why?Yes I think the answer of your question is $2^{n-5}$ – 2017-01-11
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    I'm too lazy to write a solution. Here's an idea you may be able to use. Fix your attention on one particular side of the $n$-gon. Draw a triangle on that side; there are $n-2$ possible choices. The triangle divides the original polygon into two smaller polygons. (In most cases; the end cases may need special consideration.) Now each of the smaller polygons must be triangulated, ***but*** one edge of the smaller polygon doesn't "count" because it's not an edge of the original polygon. – 2017-01-11
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    @bof what should we do when it divides into two polygons it will dive us again the same problem? – 2017-01-11
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    Maybe look at ... https://en.wikipedia.org/wiki/Catalan_number. – 2017-01-11

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Such a triangulation has $n-4$ triangles of type I having one edge on $\partial P$ and two triangles of type II having two edges on $\partial P$. The centers of these triangles are the vertices of a linear graph with the triangles of type II at its ends.

Begin by placing a triangle $T_0$ of type II. This can be done in $n$ ways. Then follow the $n-4$ triangles $T_i$ $(1\leq i\leq n-4)$ of type I, whereby each of them shares one edge with $T_{i-1}$, and we can choose whether the edge on $\partial P$ shall be on the left or on the right. The last triangle $T_{n-3}$, of type II, is then already there.

So far we have $n\cdot 2^{n-4}$ choices, but each configuration is counted twice since we might have begun as well with $T_{n-3}$ instead of $T_0$. The end result therefore is $n\cdot 2^{n-5}$ $(n\geq4)$.