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 $\displaystyle 180^o$ minus its corresponding interior angle. That way, if any interior angle is greater than $\displaystyle 180^o$ (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 $\displaystyle 360^o$ as usual. Here's why:

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

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

$\displaystyle =720^o-360^o$

$\displaystyle =360^o$