What's a proof that the angles of a triangle add up to 180°?

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).

According to the Gauss-Bonnet theorem, if $T$ is your triangle, $\gamma_i$ its sides and $v_i$ its vertices, $$\int_T K+\sum_i\int_{\gamma_i}\kappa + \sum_i\alpha_i=2\pi\chi(T)$$ with $K$ the Gaussian curvature, $\kappa$ the geodesic curvature along the sides, $\alpha_i$ the external angle at the vertex $v_i$ (measured in radians), and $\chi(T)$ the Euler characteristic of $T$. Since the plane is flat, $K\equiv 0$; since the sides of the triangle are geodesics, $\kappa\equiv0$; and since $T$ is contractible, $\chi(T)=1$. Therefore the above formula tells us that $$\sum_i\alpha_i=2\pi.$$ Since the internal angle at the $i$th vertex is $\pi-\alpha_i$, this tells us that the sum of the internal angles is $\pi$, which is what we wanted

There is always the "taxicab" proof. Start at a vertex and take a cab trip around the triangle. The cab will have turned through an angle of $360^{\text{o}}$ Extend the sides of the triangle appropriately and you will get three $180^{\text{o}}$ angles with the vertex angles being supplementary to the turns taken by the cab. Then note that $540-360=180$ ...

I know this isn't a direct answer, but it is my favourite demonstration.