Prove that if all triangles have the same angle sum then the sum of the angles in any triangle must be 180. I don't know where to start. 
I know that the sum of the angles is less than or equal to 180. 
but how do i prove this.
 A: Let the sum be $x$. Cut an arbitrary triangle into two triangles. The sum of the smaller triangles is $2x$ but also $x+180^\circ$.
A: It seems what must have been intended is this: Find one triangle in which it is easy to prove that the sum is $180^\circ$.
Then deduce that if it's the same in all other triangles, then it must be $180^\circ$ in all triangles.
PS: OK, let's look at the isosceles right triangle.  The conventional way to view it would say that since one angle is $90^\circ$ and the other two are equal to each other, and the sum of all three must be $180^\circ$, it follows that the sum of those two must be $90^\circ$, so each must be $45^\circ$.  But we cannot do that here because the fact that the sum is $180^\circ$ is what we must prove rather than a fact that we can rely on in this context.  So here's what I propose: Draw the diagonal of a square, dividing it into two triangles of the shape just described.  The diagonal splits the $90^\circ$ angle at one corner of the square into two parts.  Argue that those two parts are congruent, so each must be half of $90^\circ$.  Then you have the two $45^\circ$ angles, so the sum is proved to be $180^\circ$.
Finally, if the sum must be the same in all triangles, then $180^\circ$ is it.
A: Draw a generic triangle. Produce a line through a vertex parallel to the opposite side. Applying alternate angles, the result is seen. 
