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  1. #1
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    Question 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
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  2. #2
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    Hello hebby
    Quote Originally Posted by hebby View Post
    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.

    Grandad
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    So would the exterior angles sum up to 0?
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  4. #4
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    Hello hebby
    Quote Originally Posted by hebby View Post
    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
    Grandad
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