A way to directly see that the interior angles of triangle sum to $180^\circ$? I'm looking for a way to look at a triangle, and perhaps visualize a few extra lines, and be able to see that the interior angles sum to $180^\circ$.

I can visualize that supplementary angles sum to $180^\circ$. I'd like to be able to see the interior angle sum similarly...

I can see that the exterior angles must sum to $360^\circ$, because if you walked around the perimeter, you would turn around exactly once (though I can tell this is true, I don't really see it). I also saw a proof on KA, where the exterior angles were superimposed, to show they summed to $360^{\circ}$ (though I'm not 100% comfortable with this one).

Finally, for $a$, $b$, and $c$ exterior angles $a+b+c=360$:

\begin{align}
 (180-a) + (180-b) + (180-c) &amp; = 3\times 180 - (a+b+c) \\
                             &amp; = 3\times 180 - 360 \\
                             &amp; = 180 \\
\end{align}

But I find this algebra hard to see visually/geometrically. Is there proof that enables one to directly see that the interior angles of triangle sum to $180^\circ$?

A couple of secondary questions:


am I visually deficient in my ability to imagine?
or, am I asking too much of a proof, that I be able to see it, and that beimg able to tell that it is true should be enough...?
Since that fact about the angle sum is equivalent to the parallel postulate, any visualization is likely to include a pair of parallel lines. Here's one from wikipedia:
Imagine a triangle $ABC$ with two of the sides, say $AB$ and $BC$, rigid but meeting at the joint $B$, which is movable in the plane of the triangle, like a hinge sort of (also, you may fix one of these rigid sides, say $BC$, and rotate the other). Let the side $AC$ be made of a sufficiently elastic material (at least so that $AC\leq AB+BC$ ).

If you play with this imaginary object, you should soon see that as you increase $A \hat B C$, the other two angles eventually diminish (certainly for $A \hat B C&gt;π/2$). Indeed, you may extend $\angle ABC$ as wide as you please close to $\pi$, so that the two rigid sides approach a single rectilinear segment; then the angle between them becomes arbitrarily close to $π$ and the other two angles are vanishing; in the limiting (degenerate) position the side $AC$ coincides with the segment formed by the rigid sides. 

If you can already see that any point in this sequence of transformations of $\triangle ABC$ the sum of angles is conserved, then it is clear why for any triangle this sum is always $π$.