What's a proof that the angles of a triangle add up to 180°? Back in grade school, I had a solution involving "folding the triangle" into a rectangle half the area, and seeing that all the angles met at a point:



However, now that I'm in university, I'm not convinced that this proof is the best one (although it's still my favourite non-rigorous demonstration). Is there a proof in, say, linear algebra, that the sum of the angles of a triangle is 180 degrees? Or any other Euclidean proofs that I'm not aware of?
Here's a decent Euclidean proof:

Let $x$ be the line parallel to side $AB$ of $\triangle ABC$ that goes through point $C$ (the line is unique because of the fifth postulate). $AC$ cuts $x$ and $AB$ at the same angle, $\angle BAC$ (corollary of the fifth postulate). $BC$ cuts $x$ and $AB$ at the same angle, $\angle ABC$. These two angles and the final angle $\angle ACB$ form a straight angle on $x$, which is always  $180^\circ$ (corollary of the third postulate).
If we have a Triangle with vertices A,B,C 

As we can see in the Triangle, if a man starts from Vertices A and goes in the direction of the line extended from Vertices A, suddenly decides to go towards Vertices B, Meaning he has to turn in Anti-Clockwise Direction for Angle(PI - A). now he walks to Vertices B, then suddenly he wants to go towards C and turns  in Anti Clockwise Direction for Angle(PI - B) and moves along C, finally he does the same and turn towards A with a Rotation of Angle(PI - C), So completes One Revolution or 2*PI Radians, or the sum of all the Three Rotations the man took to complete one Perimeter walk on the Triangle Walk or an equivalent of Angle(3*PI -(A+B+C)).

Meaning we Reach an Equation :
Angle(2*PI) = Angle(3*PI -(A+B+C))
=> Angle(A+B+C) = Angle(2*PI)

Assumptions Taken to Calculate This result :


A rotation of Angle(PI) is required to move in an exactly opposite direction.
Obviously to come to the same line of sight we need to rotate an angle of Angle(2*PI).


I think i can safely assume these and get to the result.