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I'm trying to calculate the length of the edge [B'A] of a tetrahedron (if possible):

tetrahedron

(Direct link)

I know the lengths [DA] and [BA] (and consequently [DB]), and every face is a right-angled triangle (but not in such a way as to form a regular tetrahedron).

Intuitively, I feel the triangle B'DB is "projected" onto B'AB, and that in Thales theorem ish way:

$\frac{[BD]}{[BA]} = \frac{[B'D]}{[B'A]}$

But I'm not sure where to go from here. I can express the length using trig / Pythagores, but I end up with a system of equations with more variables than equations - I'm having a hard time expressing all the constraints the right angles add.

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  • $\begingroup$ The picture doesn't open, perhaps you might link it via another site. (There were sites where you post a picture and put a link on it so other people can see, you probably know it better than me) $\endgroup$ – Deniz Tuna Yalçın Oct 9 '17 at 15:27
  • $\begingroup$ @DenizTunaYalçın Apologies, I added a link below the image $\endgroup$ – rthur Oct 9 '17 at 15:32
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Your data are not enough to determine $AB'$. Point $B'$ could be anywhere on the circle of diameter $BD$ in the plane perpendicular to $AD$. Hence $AB'$ can take any value between $AD$ and $AB$.

enter image description here

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  • $\begingroup$ Thanks for the figure, that made it very clear! $\endgroup$ – rthur Oct 9 '17 at 16:22

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