Area of a spherical polygon A spherical polygon on $S^2$ is the region formed by the intersection of $n$ hemispheres of $S^2$, where $n$ is an integer $\geq 3$. 
Show that, if $\alpha_1,\cdots,\alpha_n$ are the interior angles of such a polygon, its area is equal to $$\sum_{i=1}^n \alpha_i − (n − 2)\pi$$ 
Could you give me some hints how we can show this? 
 A: Given a spherical polygon with $n$ sides, we can divide it in $n-2$ adjacent triangles $T_i$ with a common vertex in a vertex (say $C$) of the polygon, such that the area of the polygon $A$ is the sum of the areas $A_i$ of the triangles. 
Now, if $a_i,b_i,c_i$ are the internal angles of the triangle $Ti$, the Girard's Theorem states that $a_i+b_i+c_i=\pi+A_i$ where $A_i$ is the area of the triangle.
So, summing for the  $n-2$ triangles, we have:
$$
\sum_{i=1}^{n-2}\left( a_i+b_i+c_i\right)=(n-2)\pi+\sum_{i=1}^{n-2}A_i
$$
Now it is not difficult to  see that the sum at left is the the sum of all the internal angles of the polygon $\alpha_i$, so that we have:
$$
\sum_{i=1}^{n}\alpha_i=(n-2)\pi+ A
$$
Clearly here the key step is the proof of the Girard's Theorem, that is not difficult , but require a good figure that I'm not able to construct, so I give this link.
Since you have used the ''differential geometry'' tag, note that the Girard's Theorem is a special case of Gauss-Bonnet Theorem.

In this figure the decomposition is illustrated for a plane polygon with $n=6$, decomposed in $n-2=4$ triangles. The figure represents a plane polygon, but for a spherical one the situation is the same. Note that the sum of the internal angles of the triangles gives the sum of the angles of the polygon. A rigorous proof for any $n$ can be given by induction.

