TITLE: What's a proof that the angles of a triangle add up to 180°?
QUESTION [80 upvotes]: 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?

REPLY [5 votes]: It's a (trivial) calculation.
Background
Let $A$, $B$, and $C$ be distinct points in the plane.  Considering them as complex numbers, recall that the angle $\theta$ at a vertex like $ABC$ is the angular part $\theta$ of $(C-B)/(A-B) = r\exp(i\theta)$.  It is written $\theta = \arg(r\exp(i\theta))$ and is defined only modulo $2\pi$, although conventionally a representative in the interval $(-\pi, \pi]$ is taken.
Observe that $\arg(z w) =\arg(z) + \arg(w)$ for any nonzero complex numbers $z$ and $w$: this follows immediately from the multiplicative property of the exponential, $\exp(i\theta)\exp(i\phi) = \exp(i(\theta + \phi))$.
Proof
The sum of the three angles at vertices $A$, $B$, and $C$ is obtained merely by rearranging the product of three numbers in the denominator of a fraction:
$$\eqalign{
\arg\frac{B-A}{C-A} + \arg\frac{C-B}{A-B}+\arg\frac{A-C}{B-C} &= \arg\left(\frac{B-A}{C-A}\frac{C-B}{A-B}\frac{A-C}{B-C}\right) \\
&= \arg\left(\frac{B-A}{A-B}\frac{C-B}{B-C}\frac{A-C}{C-A}\right) \\
&= \arg\left(-1^3\right) = \arg(-1) \\
&= \pi.
}$$
Because the points are distinct, all three angles are strictly between $-\pi$ and $\pi$. The only way three of them can sum to $\pi$ (modulo $2\pi$) is for the sum actually to equal $-\pi$ or $\pi$, QED.

Comments
Both sums, $-\pi$ and $\pi$, are possible: in the former case, the triangle $ABC$ is traversed in the negative direction (with its interior to the right) while in the latter it is traversed positively.
This argument does not generalize (without taking special care) to four or more points, because the usual result found in elementary geometry expositions is not generally true (namely, that the sum of interior angles of an $n$-gon is $(n-2)\pi$).  E.g., with four points we might care to show that the sum of angles in a 4-gon (a "shape with four sides") is $2\pi$.  Consider the points at $1$, $i$, $-i$, and $-1$: in that order.  Each of the angles, considered in isolation, is $\pi/4$, so their sum is $\pi$. We need to use correctly oriented angles, however; in traversing this 4-gon, two of the angles will equal $\pi/4$ and two will equal $-\pi/4$, summing to $0$.  This is correct: the result is congruent to $2\pi$ modulo $2\pi$, as it should be, but it is not equal to $\pm 2\pi$.