planarity of a quadrilateral If I have a quadrilateral in $\mathbb{R}^3$, is it true that the quadrilateral is simple and planar if and only if the interior angles sum to $2\pi$?
I'm sure this fact is very elementary, but some searching didn't turn up a proof.
 A: The key is spherical geometry's Triangle Inequality: Given three rays emanating from a point, any one of the angles formed is less than, or equal to, the sum of the other two (and the "equal to" case corresponds to the rays being coplanar).
To apply this fact to your question, let $ABCD$ be the quadrilateral, which we decompose into triangle by drawing diagonal $AC$. Then we manipulate the angle-sums of those triangles:
$$\begin{align}
2\pi &= (\angle B + \angle BAC + \angle BCA) + (\angle D + \angle DAC + \angle DCA)\\
&= (\angle BAC + \angle DAC) + \angle B + (\angle BCA + \angle DCA) + \angle D \\
&\ge \phantom{(\angle BAC}\angle A\phantom{\angle DAC} + \angle B + \phantom{(\angle BCA}\angle C\phantom{\angle DCA} + \angle D
\end{align}$$
We get $\angle A = \angle BAC + \angle DAC$ when, and only when, $\overrightarrow{AB}$, $\overrightarrow{AC}$, $\overrightarrow{AD}$ are coplanar (similarly for $\angle C = \angle BCA + \angle DCA$), which is to say that the quadrilateral itself is coplanar. Done.

I'll point out that the sophisticated version of the Triangle Inequality is the Law of Cosines; for instance, writing $\theta$ for the dihedral angle along $AC$, the angles at vertex $A$ satisfy ...
$$\cos\angle A = \cos\angle BAC \cos\angle DAC + \sin\angle BAC \sin\angle DAC \cos\theta$$
We get the Triangle Inequality by observing that $-1 \le \cos\theta \le 1$ implies
$$\cos(\angle BAC + \angle DAC) \le \cos\angle A \le \cos(\angle BAC - \angle DAC)$$
so that 
$$|\angle BAC - \angle DAC| \le \angle A \le \angle BAC + \angle DAC$$
Note that equality occurs for $\cos\theta = \pm 1$, whereupon the triangles on either side of $AC$ are "opened" or "folded" flat: they're coplanar.
In Euclidean geometry, the same thinking takes us from
$$c^2 = a^2 + b^2 - 2 a b \cos C$$
to 
$$(a-b)^2 \le c^2 \le (a+b)^2$$
to the Euclidean Triangle Inequality:
$$|a-b| \le c \le a+b$$
A: Without restriction of generality you may assume two consecutive vertices at $x_0:=(0,0,0)$ and $x_1:=(1,0,0)$. Then $x_2=(1,a,b)$ with $a^2+b^2>0$ and $x_3=(0,c,d)$ with $c^2+d^2>0$. It follows that $x_2-x_3=(1,a-c,b-d)$ and therefore
$$0=(x_3-x_0)\cdot(x_2-x_3)=c(a-c)+d(b-d)\ ;$$
and similarly
$$0=(x_2-x_1)\cdot(x_2-x_3)=a(a-c)+b(b-d)\ .$$
Subtracting the last two equations we obtain
$$(a-c)^2+(b-d)^2=0\ ,$$
which implies $a=c$ and $b=d$. It follows that our four points are $(0,0,0)$, $(1,0,0)$, $(1,a,b)$, and $(0,a,b)$; so they are the vertices of a planar rectangle.
