2
$\begingroup$

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 ?

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

enter image description here

$\endgroup$
  • $\begingroup$ Thanks. Why $DC > DH$ ? you did not define $N$. $\endgroup$ – Julien Narboux Jul 24 '17 at 10:07
  • $\begingroup$ You are right: I forgot to write that $N$ is the projection of $C$ onto $BD$. We have $DC>DH$ because $DCH$ is a right-angled triangle and $DC$ is its hypotenuse. $\endgroup$ – Aretino Jul 24 '17 at 10:36
  • $\begingroup$ Why DCH is right angles triangle ? $\endgroup$ – Julien Narboux Jul 24 '17 at 11:18
  • $\begingroup$ Planes $\beta$ and $\delta$ are perpendicular to $\alpha$. Their intersection $a$ is thus also perpendicular to $\alpha$. $\endgroup$ – Aretino Jul 24 '17 at 13:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.