exterior angle of a reflex angle How are exterior angles for reflex angles defined? Are they negative? This doubt came to my mind while thinking about the sum of the exterior angles of a concave polygon. I think it should be negative because this would allow for the sum of the exterior angles to add up to 360 degrees, just like in the convex polygon counterpart.
 A: I worry about Joseph's language (turn angle and external angle) and maybe it's different to UK. We use interior and exterior only. Interior is self explanatory- sum of interior = (n-2)x pi rad/180 degrees. Interior + Exterior = pi rad/180 degrees at each vertex, ie is the angle made by extending a side at one vertex. Sum of exterior angles =2pi rad/360 degrees.
As you say, Apekshik, this must mean that the exterior angle of reflex angles is negative. If you remember the derivation of the rule (form triangles at the centre of the shape, find angle at base, then find exterior by angles on a straight line- works best for regular shapes but can be done for any polygon- angles at centre sum to 2pi/360 so the exterior angles do as well), what do the triangles look like and where are the exterior angles? Pretty strange because the exterior angles are inside the shape (and negative in this sense). So that the red, exterior, angle is negative in the sense that is inside not outside the shape. I add the beginning of the proof for sum of exteriors but each exterior is half of two central angles and when summed they make the sum of the central angles.

A: I think you are mixing up the exterior angle, which is just $2\pi$ minus the
interior angle (as I tried to explain in your previous question 
here),
and the turn angle at a vertex.
Orient the edges of the polygon to be counterclockwise (ccw). Imagine walking
along an edge. The amount you need to turn to walk along the next edge
is the turn angle at that vertex. For convex vertices, you turn left, ccw, positive. For reflex vertices, you turn right, cw, negative. And the sum of
the turn angles is always $2\pi = 360^\circ$, because you eventually walk
full-circle.
Whereas the sum of the external angles depends on the number of vertices
(as does the sum of the internal angles).
