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.


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.

  • $\begingroup$ 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

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.