A synthetic proof that a figure with 4 right angles is planar Assume that we have a skew quadrilateral ABCD, such that the four angles are right angles (a rectangle).
How can I prove synthetically that ABCD are coplanar ?
 A: Draw the diagonal $AC$ to obtain two right triangles. Since $\angle BAC+\angle CAD+\angle DCA+\angle ACB=\pi$, one of $\angle BAC+\angle CAD\le \frac \pi 2$, $\angle DCA+\angle ACB\le\frac\pi2$ must hold.
A: Let $M$ and $N$ be the midpoints of the diagonals $[AC]$ and $[BD]$. If $M=N$ then the quadrangle  is a planar rectangle. If $M\ne N$ then the  two spheres $S_M$ with center $M$ and diameter $[AC]$, resp., $S_N$ with center $N$ and diameter $[BD]$ are different, but both contain all four points $A$, $B$, $C$, $D$. It follows that all four points lie on the circle $S_M\cap S_N$, hence form a planar rectangle, contrary to our assumption.
A: Consider plane $\alpha$ of triangle $ABD$ (where $BD$ is a diagonal of $ABCD$). Line $CD$ is perpendicular to $AD$, thus $CD$ belongs to plane $\delta$, passing through $D$ and perpendicular to $AD$. For analogous reasons, $BC$ belongs to plane $\beta$, passing through $B$ and perpendicular to $AB$. It follows that $C$ belongs to line $a$, which is the intersection of $\delta$ and $\beta$.
Let $H$ be the intersection of plane $\alpha$ with $a$: from plane geometry we get $\angle DHB=90°$. Suppose now, by contradiction, that $C$, on line $a$, is different from $H$. In that case $DC>DH$ and $BC>BH$, so that triangle $BCD$ has a side in common with triangle $BHD$ but the other two sides are greater. 
Rotate now triangle $BCD$ around $BD$, so that rotated triangle $BC'D$ lies on plane $\alpha$. If $N$ is the projection of $C$ onto $BD$, from the external angle theorem we have $\angle DC'N<\angle DHN$ and $\angle BC'N<\angle BHN$. The feet of the altitude of a right-angle triangle lies inside the opposite side so we can add angles to get $\angle DC'B<\angle DHB$. We have then $\angle BCD=\angle BC'D<90°$, against the hypothesis.

