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