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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 $a,b,c,d$.

enter image description here

Now what I did is the following: enter image description here

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.

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2 Answers 2

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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.

Relocated exterior angle cº

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

We see that

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

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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.

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