Four points in $\Bbb R^3$ form equal angles then |AB|=|CD| and |BC|=|DA|.

Question

I'm interested in the following problem:

If $A,B,C,D$ are four distinct points in $\Bbb R^3$ satisfying $\angle ABC=\angle BCD=\angle CDA=\angle DAB$,
then is it must be true that $|AB|=|CD|,|BC|=|DA|$?

If $A,B,C,D$ are coplanar then $ABCD$ is a rectangle, so $|AB|=|CD|,|BC|=|DA|$.
If $A,B,C,D$ are not coplanar then I don't know how to prove.

For $|AB|=|BC|=|CD|=|DA|$ I have an example:
$A = (0, -1, -a)$
$B = (-1, 0, a)$
$C = (0, 1, -a)$
$D = (1, 0, a)$

I can rephrase the problem in terms of vectors $B-A,C-B,D-C,A-D$:

If four vectors $\vec a,\vec b,\vec c,\vec d$ in $\Bbb R^3$ satisfy$$\vec a+\vec b+\vec c+\vec d=\vec0$$and$$\frac{\vec a\cdot\vec b}{|\vec a||\vec b|}=\dots=\frac{\vec d\cdot\vec a}{|\vec d||\vec a|}\tag1$$then is it must be true that $|\vec a|=|\vec c|,|\vec b|=|\vec d|$?

I can write the equation $(1)$ in components $$\frac{a_1 b_1+a_2 b_2+a_3 b_3}{\sqrt{a_1^2+a_2^2+a_3^2} \sqrt{b_1^2+b_2^2+b_3^2}}=\frac{b_1 c_1+b_2 c_2+b_3 c_3}{\sqrt{b_1^2+b_2^2+b_3^2} \sqrt{c_1^2+c_2^2+c_3^2}}=\frac{c_1 d_1+c_2 d_2+c_3 d_3}{\sqrt{c_1^2+c_2^2+c_3^2} \sqrt{d_1^2+d_2^2+d_3^2}}=\frac{a_1 d_1+a_2 d_2+a_3 d_3}{\sqrt{a_1^2+a_2^2+a_3^2} \sqrt{d_1^2+d_2^2+d_3^2}}$$ then I square the equation $$\frac{\left(a_1 b_1+a_2 b_2+a_3 b_3\right){}^2}{\left(a_1^2+a_2^2+a_3^2\right) \left(b_1^2+b_2^2+b_3^2\right)}=\frac{\left(b_1 c_1+b_2 c_2+b_3 c_3\right){}^2}{\left(b_1^2+b_2^2+b_3^2\right) \left(c_1^2+c_2^2+c_3^2\right)}=\frac{\left(c_1 d_1+c_2 d_2+c_3 d_3\right){}^2}{\left(c_1^2+c_2^2+c_3^2\right) \left(d_1^2+d_2^2+d_3^2\right)}=\frac{\left(a_1 d_1+a_2 d_2+a_3 d_3\right){}^2}{\left(a_1^2+a_2^2+a_3^2\right) \left(d_1^2+d_2^2+d_3^2\right)}$$ then I am stuck.

Answer 1

If we define the lengths of line segments as follows:

$ a = |AB|, b = |BC| , c = |CD| , d = | DA | $

Then, from the law of cosines,

$|AC|^2 = a^2 + b^2 - 2 a b cos \theta = c^2 + d^2 - 2 c d \cos \theta $

and

$ |BD|^2 = a^2 + d^2 - 2 a d \cos \theta = b^2 + c^2 - 2 b c \cos \theta $

Fix $\theta, a$ and $ b$ to some values, and solve for $c$ and $d$.

I fixed $\theta = \dfrac{\pi}{6} $ , $a = 1 , b = 1.2 $ and got the following solutions for $c,d$

$ (c, d) = (1, 1.2) , (1, 0.532050808), (1.078460969, 1.2) $

So we don't necessarily have to have $ |AB| = |CD|$ and $| BC | = |DA| $.

Answer 2

Unless I misunderstand, here's a quick visual disproof:

You may extend the lines BC and AD arbitrarily (but by the same distance) to make DC whatever length you like without changing any of the interior angles.