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Originally Posted by OngBonga
Here's a visualisation to make it really clear.
Yeah, there's an argument in the comments that is helpful and elegant.
If you put 2 of the square base pyramids side by side, sharing an edge of their base,
then it's clear that the appropriate sides of those pyramids are coplanar, by construction.
(The sides have bases which are colinear, and their vertices imply a line parallel to their base.)
If you construct that line connecting the peaks of the pyramids, you construct the regular tetrahedron*,
which, since it shares all vertices with 2 coplanar shapes, is coplanar to those shapes, too.
*We know 2 of its faces are equilateral triangles, since they mate perfectly to the "inner" sides of the pyramids,
and we know that the distance between the 2 peaks of the pyramids is also equal to the edge lengths of all the edges under discussion.
So by extension, we've proved that all the edge lengths of the constructed tetrahedron are equal, and thus it is a regular tetrahedron.
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