How do we define exterior angle of a concave polygon whose interior is reflex?

Question

How do we define exterior angle of a concave polygon whose interior is reflex?

I have seen in few books and websites saying that, sum of all exterior angles of a concave polygon is $360$ degrees.

Exterior angle definition in the case of concave polygons.

I saw the post above, which says that exterior of reflex interior is $360^{\circ}$ minus that. In that case I took an example of a concave quadrilateral shown below with exterior angles as $a,b,c,d$.

Now what I did is the following:

We see that $$x+180^{\circ}-a=c$$ and also $$180^{\circ}-x+180^{\circ}-b=d$$

Adding both, we get $$a+b+c+d=540^{\circ} \ne 360^{\circ}$$

Please give some inputs on this.

Answer 1

my friend please note that the sum of the exterior angle is not equal to the sum of the interior angle in any polygon. What you found is partially the sum of exterior angles, If you add 180+a+c+180+b+180+d=540+a+b+c+d=sum of exterior angle (which is not equal to sum of interior angle). your idea is very good but it leads to mischievous false proof. lets say 180-a+360-c+180-b+180-d=360 we get 900=a+b+c+d+360 since a+b+c+d =540, 900-540=360. that was where the mistake was.

Answer 2

Your $c^\circ$ is not an exterior angle of the concave quadrilateral (the kind of exterior angles that add to $360^\circ$).

As illustrated on Wikipedia, the exterior angle at that reflex vertex is negative, formed by one side and a line extended from the adjacent side.

Let $c^\circ$ be the (signed) exterior angle, which is negative. Its magnitude is $\lvert c\rvert^\circ = -c^\circ$.

We see that

$$\begin{align*} x^\circ &= \left(180^\circ - b^\circ\right) + \left(180^\circ - d^\circ\right)\\ &= 360^\circ - b^\circ - d^\circ\\ a^\circ &= x^\circ + \lvert c\rvert^\circ\\ &= 360^\circ - b^\circ - d^\circ - c^\circ\\ a+b+c+d &= 360 \end{align*}$$