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If a closed polygonal chain of length 4 forms only right angles, is it necessary that the 4 points are coplanar and hence that the chain forms a rectangle? Is it still the case if it is only known for 3 of the angles?

A counterexample could be a tetrahedron with faces being right-triangles with the right angle at a different vertex each.

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Let $ABC$ be a right triangle with hypothenuse $AC$, suppose that $\angle BCD$, $\angle CDA$, $\angle DAB$ are right. Let $D'$ be the projection of $D$ onto the plane of $ABC$, so that $ABCD'$ is a rectangle. By repeated use of Pythagoras, $$(CD'^2+DD'^2)+(D'A^2+DD'^2)=CD^2+DA^2=AC^2=AB^2+BC^2.$$ Since $AB=CD'$ and $BC=D'A$, we can cancel them out to conclude $DD'=0$ and thus $D=D'$, meaning that $ABCD$ is a rectangle.

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  • $\begingroup$ Why is $ABCD'$ a rectangle? $\endgroup$
    – Joce
    Apr 15, 2023 at 12:27
  • $\begingroup$ If $\angle BCD$ is right, so is $\angle BCD'$, likewise for $\angle D'AB$. $\endgroup$
    – ViHdzP
    Apr 15, 2023 at 20:43

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