Can a quadrilateral with four not coplanar points have four right angles?

Question

Can all of its angles be 90°? My guess is yes, but I can't properly prove it

Answer 1

Let $ABCD$ your "quadrilireral".

Thus, $\measuredangle ABD=\measuredangle BCD=90^{\circ}$, but $AB$ and $CD$ are not coplanar.

Now, there are unique planes $\alpha$ and $\beta$ such that $\alpha||\beta$, $AB\subset \alpha$ and $CD\subset \beta$.

Now, if $\measuredangle BAD=\measuredangle ADC=90^{\circ}$ then $AD\perp \alpha$, which says $AD||BC$, which is a contradiction.

Answer 2

Consider the quadrilateral $ABCD$ with four right-angles. The diagonal $AC$ subtends angles of $90^{\circ}$ at $B$ and $D$ so the points lie on a sphere centre the midpoint of $AC$. Similarly they lie on a sphere centre the midpoint of $BD$, whence the midpoints coincide.

Thus the diagonals bisect each other, and in particular lie in the same plane, whence their endpoints, the vertices of the quadrilateral, lie in the same plane.

Answer 3

Its simply not possible. Consider $ABC$ figure where $AB = BC = a$ and $AB \perp BC$. Then we draw two circles perpendicular to plane of $ABC$ centred at $A$ and $C$ of radius $a$ (side length of square). These circles represent possible position for point $D$ of square.

Then we see these circles intersect only once! That too in plane of $ABC$. So here lies the fourth point, in plane of $ABC$