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Let $ABCD $ a tetrahedron s.t. $ \angle ABD=\angle BDC $ and $ \angle BAC=\angle ACD$.

Show that $AB=CD $.

I construct $DD_1, CC_1 \perp AB $, $C_1, D1\in AB $ and $AA_1, BB_1\perp CD $ , $B_1, A_1\in CD $. By congruences of triangles we have $BD_1=DB_1$ and $AC_1=CA_1$. If $AB\perp CD$ then it is easy because $C_1=D_1$ and $A_1=B_1$.

Now I am stuck.

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After several failed attempts to disprove this, I've finally got it.

Let $\alpha,\beta,\gamma,\delta$ be the vectors $B-A,D-B,C-D,A-C$ respectively. Since $A,B,C,D$ don't lie in a plane, no two of these vectors are parallel.

Choose coordinates so that $\alpha=\|\alpha\|\cdot (\cos\theta,\sin\theta,0)$ and $\gamma=\|\gamma\|\cdot (\cos\theta,-\sin\theta,0)$; the first coordinate axis is parallel to $\frac{\alpha}{\|\alpha\|}+\frac{\gamma}{\|\gamma\|}$, the second is parallel to $\frac{\alpha}{\|\alpha\|}-\frac{\gamma}{\|\gamma\|}$, and the third is orthogonal to both $\alpha$ and $\gamma$. Since $\alpha$ and $\gamma$ are not parallel, $\sin\theta\neq 0$. What's the point? Let $v=(x,y,z)$ be an arbitrary vector. We have $$\frac{\langle v,\alpha\rangle}{\|v\|\cdot\|\alpha\|}=\frac{x\cos\theta+y\sin\theta}{\sqrt{x^2+y^2+z^2}}\text{ and }\frac{\langle v,\gamma\rangle}{\|v\|\cdot\|\gamma\|}=\frac{x\cos\theta-y\sin\theta}{\sqrt{x^2+y^2+z^2}}$$ These are equal if and only if $y=0$. We have a nice characterization of all vectors that make equal angles with $\alpha$ and $\gamma$.

Now, we look at the conditions on our triangles. $\angle ABD$ is the angle between $\alpha$ and $\beta$, $\angle BDC$ is the angle between $\beta$ and $\gamma$, $\angle BAC$ is the angle between $\delta$ and $\alpha$, and $\angle ACD$ is the angle between $\gamma$ and $\delta$. The first two being equal means that $\beta$ makes equal angles with $\alpha$ and $\gamma$. The last two being equal means that $\delta$ makes equal angles with $\alpha$ and $\gamma$. Thus both $\beta$ and $\delta$ lie in that plane $y=0$.

Finally, $\alpha+\beta+\gamma+\delta=B-A+D-B+C-D+A-C=0$. Look at the $y$-coordinate of this sum $\alpha+\beta+\gamma+\delta$: $$0 = \|\alpha\|\cdot\sin\theta+0+\|\gamma\|\cdot(-\sin\theta)+0=(\|\alpha\|-\|\gamma\|)\sin\theta$$ As previously noted, $\sin\theta\neq 0$ and we must have $\|\alpha\|=\|\gamma\|$. Those are the lengths $AB$ and $CD$, and we're done.

On a side note, the three-dimensional nature of this is crucial. The claim is false for four points $A,B,C,D$ in a plane, as we could have a trapezoid $ABCD$ with $AB$ and $CD$ parallel.

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  • $\begingroup$ The claim is true for $A,$ $B$, $C$, $D$ in a plane as it would force an isosceles trapezium with $AC$ and $BD$ parallel. $\endgroup$ – Anubhab Ghosal Feb 14 '19 at 9:29
  • $\begingroup$ No, it's configuration-dependent. In the configuration I'm thinking of, segments $AC$ and $BD$ intersect. Also note - I removed "isosceles", as it's unnecessary. $\endgroup$ – jmerry Feb 14 '19 at 9:39
  • $\begingroup$ That is true. It depends on the configuration. $\endgroup$ – Anubhab Ghosal Feb 14 '19 at 9:58

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