anglegeometrypolygonsquadrilateraltriangles
Then I tried to find the sum of interior angles of some concave polygons. Surprisingly, it seems that the formula above also applies.
Can anyone help me prove this?
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
Since that fact about the angle sum is equivalent to the parallel postulate, any visualization is likely to include a pair of parallel lines. Here's one from wikipedia:
Best Answer
Take a point at the center of the convex polygon. For each side of the polygon, form a triangle by taking two lines from the center point to the endpoints of the side. Those two lines and the side will form a triangle. Since there is one triangle per side, this will give a total of $n(180^{\circ})$. To get the sum of the internal angles we subtract the angle between the two lines emanating from the center point from each triangle. But the sum of the angles from the center point is $360^{\circ}=2(180^{\circ})$.
Therefore, the sum of the internal angles is $n(180^{\circ})-2(180^{\circ})=(n-2)180^{\circ}$