Why sum of interior angles in convex polygon is $(n-2)\cdot 180$ couple days ago in my math high school lessons I learned that sum of interior angles in convex polygon is:
$Z$ = sum of the angles, $n$ = number of sides in polygon
$Z=(n-2)\cdot 180$
Can someone help me understanding this formula, and why is it like this?
 A: Start with a triangle. 
Choose an edge on the triangle and mark a point on it. 
Now imagine pulling the point outwardly away from the edge. 
What happens is the original triangle gains 2 more edges and there is an extra $180^{\circ} $ from the triangle formed by the original edge and the two new ones.
This has the advantage of explaining why there are $n-2$ less triangles as each triangle is paired with the $2$ parts of a broken edge.
If I get chance I'll add an animation or if someone wants to edit feel free. This can be extended to make the polygon grow.
A: See the case, when $n=3$ i.e. the polygon is triangle. Now sum of all interior angles of a triangle is $180^0$ which matches with formula.
For $n>3$, we can divide the polygon into $n-2$ triangles. So, the sum of interior angles will be $(n-2)180$.
A: Here is a useful diagram:

The sum of all the angles in all the triangles equals the sum of the interior angles of the polygon. Notice that the shape has $7$ sides, and we are able to fit $5$ triangles inside it each of whose angles sum to $180$ degrees. This generalises to a $n$ sided polygon being able to fit $n-2$ triangles inside it as shown above. Then the sum of the angles must = $(n-2)*180$ degrees
A: Hint Pick a point in the interior of the polygon. Then, drawing line segments from the point to each vertex divides the polygon into triangles, but you already know that the sum of the angles of a triangle is $180^{\circ}$.
A: One way to understand this is to look at the exterior angles in a traverse one way around the polygon - the exterior angle being the angle turned in passing each vertex. (For example, in a regular decagon, the exterior angle at each vertex is $36°$.) Clearly the sum of such angles $\{\epsilon_i\}$ is $360°$, since you arrive back on the starting edge having turned fully around the polygon. The interior angles $\{\theta_i\}$ are each $\theta_k=180°-\epsilon_k$, so you have 
$$\sum^n \theta_i = \sum^n (180°-\epsilon_i) = n\cdot180° -  \sum^n \epsilon_i =  n \cdot 180°-360° = (n-2)180°$$

diagram for clarity

