Particular case of Gauss-Bonnet for polygons in $\mathbb{R}^2, \mathbb{H}^2$ and $\mathbb{S}^2$ (Exercise 5.16, Bonahon, Low-dimensional geometry) Let $X$ be a bounded polygon in $\mathbb{R}^2, \mathbb{H}^2$ or $\mathbb{S}^2$ consisting of finitely many disjoint convex polygons $X_1, X_2,\dotsc, X_m$. Let $\bar{X}$ be obtained by gluing together pairs of edges of $X$.
We assume that for every vertex $P$ of $X$, the angles of $X$ at the vertices that are glued together to $P$ add up to $2\pi$, so that $\bar{X}$ is a euclidean, hyperbolic, or spherical surface.
The quotient space $\bar{X}$ is now decomposed into $m$ images of the convex polygons $X_i$, $n$ images of the $2n$ edges of $X$ and $p$ point images of the vertices of $X$.
The Euler characteristic of $X$ is defined as the integer 
$$\chi(\bar{X})=m-n+p$$
I have shown the second part of the question which states: For each of the convex polygons $X_i$, $$\sum_{j=1}^{v_i} \theta_j = (v_i-2)\pi + K \text{Area}(X_i),$$ where $X_i$ has $v_i$ vertices, and $\theta_j$ are the angles of $X_i$ at the vertices. For the euclidean case we mean the usual are and $K=0$, for the hyperbolic case we have in mind the hyperpolic area and $K=-1$, whereas for the spherical case $K=1$ and Area$(X_i)$ denotes the spherical area. 
We shall use the above result in order to show that $$2\pi \chi(\bar{X}) = K \text{Area}(X)\enspace (*)$$
I have tried to write down what I know, namely $K\text{Area}(X)= K\sum_{i=1}^{m} \text{Area}(X_i)= (\sum_k \theta_k)[(\sum_i v_i)-2m]\pi $, where $k$ ranges over all the vertices in $X$ so that we sum all the angles of the polygons in $X$ and 
$\sum_i v_i$ represents the sum of all the vertices in $X$. The term $-2\pi m$  in the latter expression and the term $2\pi m$ of the left hand side in $(*)$ differ by a sign, so this may be a good start. Do you have any suggestions how to proceed? 
 A: Hint (assuming you use the correction in my comment above): You have not yet taken advantage of one of the key pieces of information given in the problem: "for every vertex $P$ of $\bar X$, the angles of $X$ at the vertices that are glued together to $P$ add up to $2\pi$". Each such set of vertices of $X$, one per vertex $P$ of $\bar X$, is called a vertex cycle of the polygon $X$. 
The general strategy here is that you want to rearrange your sum of angles $\sum \theta_j$ (where the sum is taken over all vertices of the polygon $X$) into a sum of terms with each term being a sum over a vertex cycle, hence the number of such terms equals $p$, each individual term sums to $2\pi$, and so $\sum \theta_j = 2 \pi p$. This should simplify the problem quite a bit. 
A: Following the advice of Lee Mosher: So when I rewrite $\text{Area}(X)$ by using what I have shown, namely $\sum_{j=1}^{v_i} \theta_j = (v_i-2)\pi + K \text{Area}(X_i)$ for every polygon $X_i$ I obtain $$\text{Area}(X)=\sum_{i=1}^m \text{Area}(X_i) =\sum_{i=1}^m \left[  \left( \sum_{j=1}^{v_i} \theta_j\right) - (v_i -2) \pi \right] =  \sum_{i=1}^m    \sum_{j=1}^{v_i} \theta_j  - \sum_{i=1}^m v_i \pi + \sum_{i=1}^m 2\pi.$$ The first term of the sum can be rewritten by using the hypothesis that for every vertex $P$ of $X$, the angles of $X$
 at the vertices that are glued together to $P$
add up to $2\pi$. The angles at one image point of $X$ add up to $2\pi$ and since there are $p$ image points, we obtain $\sum_{i=1}^m    \sum_{j=1}^{v_i} \theta_j  = 2\pi p$. For the latter equality we also use the fact that the $2n$ edges of $X$ get glued together pair by pair so every angle $\theta_j$ is "around" one of the vertices $p$ and conversely every vertex $p$ of $\bar{X}$ has one of the angles $\theta_j$ "around" them. As for the term $\sum_{i=1}^m v_i \pi$ we observe that the number of edges in $X$, namely $2n$ equals the number of vertices in $X$, namely $\sum_{i=1}^m v_i $ since $v_i$ is the number of vertices of the polygon $X_i$. This is a purely combinatorial fact (a triangle has 3 vertices and 3 edges, a square has 4 vertices and 4 edges and so on) and leads to the equality $\sum_{i=1}^m v_i \pi = 2\pi n$. Finally we obtain 
$$\text{Area}(X) = \sum_{i=1}^m    \sum_{j=1}^{v_i} \theta_j  - \sum_{i=1}^m v_i \pi + \sum_{i=1}^m 2\pi = 2\pi (p-n+m) = 2\pi \chi(\bar{X}).$$
Now my question is: What should I do with the constant $K$ in order to complete the proof by showing $2\pi \chi(\bar{X})= K \text{Area}(X)$? What I know is that this comes from the fact that the Euler characteristic for the euclidean case is zero, positive for the spherical case and negative for the hyperbolic case, but isn't this reasoning somewhat circular since this fact also comes from what I want to show?
