Assuming that the sum of all angles of polygon does add up to a particular constant depending on the no. of sides of the polygon.
Let us assume that the constant is 'k', if at the end of the proof we are able to find a value k for an arbitrary triangle, we can assume that k exists for all the triangles.
Assume an arbitrary triangle.
By gen. convention, each angle of triangle is named as ang.A , ang.B, ang.C, hence,
(1) ang.A + ang.B + ang.C = k
Now we all know that the triangle can be divided into 3 portions by lines joining each vertex and any point inside the triangle , (example centroid). let us name it Point P.
let The angle subtended by each side on the Point P be cen.A, cen.B, cen.C. Thus,
(2) - cen.A + cen.B + cen. C = 360deg.
Also the line joining the vertex and the Point P. of triangle will divide each angle in two parts, name each as
(3) ang.A=ang.A1 + ang.A2 and so on for ang.B and ang.C
thus for each of the smaller triangles we have following equations -
(4) ang.A1 + ang.B1 + cen.C = k
(5) ang.A2 + ang.C1 + cen.B = k
(6) ang.B2 + ang.C2 + cen.A = k
(4) + (5) + (6) will imply that
ang. A + ang.B + ang.C + cen.A + cen.B + cen. C = 3k
Now using (2) we have
ang. A + ang.B + ang.C + 360deg. = 3k
But by initial assumption and hence by (1) we have
k + 360deg. = 3k
I think this simple proof should suffice.
Or rather just cut the triangle any which way you want keeping the vertices intact. Arrange them with their pointed ends together, voila, you can see the angles must add up to 180deg.
Also why go for complicated proofs when we have simple ones. I guess, because we are Mathematecians! :)