1. ## Geometry

How would you define the exterior angles of a non convex quadrilateral ABCD so that the sum of the four exterior angles is 360

2. Hello hebby
Originally Posted by hebby
How would you define the exterior angles of a non convex quadrilateral ABCD so that the sum of the four exterior angles is 360
Define the size of each exterior angle as $180^o$ minus its corresponding interior angle. That way, if any interior angle is greater than $180^o$ (which it will be at a re-entrant vertex), its corresponding exterior angle will be negative.

3. So would the exterior angles sum up to 0?

4. Hello hebby
Originally Posted by hebby
So would the exterior angles sum up to 0?
No, they'll sum to $360^o$ as usual. Here's why:

In any quadrilateral $ABCD$ (whether convex or not) the interior angles add up to $360^o$. This is because a diagonal can be drawn that divides the quadrilateral into two triangles. The sum of the angles of each triangle is $180^o$. Therefore $\angle A + \angle B+\angle C + \angle D =360^o$, where $\angle A, ...$ denote the interior angles at each vertex.

Then if we define the size of each exterior angle as $180^o$ minus the size of the interior angle, and denote the exterior angles by $\alpha, \beta, \gamma, \delta$, we have:
$\alpha+\beta+\gamma+\delta = (180^o - \angle A)+(180^o - \angle B)+(180^o - \angle C)+(180^o - \angle D)$
$=720^o-(\angle A + \angle B+\angle C + \angle D)$

$=720^o-360^o$

$=360^o$