# Show that $AB=CD$.

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.

• The claim is true for $A,$ $B$, $C$, $D$ in a plane as it would force an isosceles trapezium with $AC$ and $BD$ parallel. – Anubhab Ghosal Feb 14 '19 at 9:29
• 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. – jmerry Feb 14 '19 at 9:39
• That is true. It depends on the configuration. – Anubhab Ghosal Feb 14 '19 at 9:58