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A triangle has the congruent isoscelizers point and equal parallelians point.

I'm pretty sure that any tetrahedron has many cutting planes that will cut an equilateral triangle. Is that much true -- that any three leg solid angle has equilateral triangle cutting planes?

If so, is there a point of concurrence for 4 of these cutting planes?

If so, is there a point where 4 congruent equilateral triangles get cut?

If not that, is there a point where 4 congruent triangles gets cut?

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    If a solid angle admits an equilateral cutting plane then there is such a plane through every point within this solid angle, and if every solid angle admits such a plane then every point in the interior of a tetrahedron belongs to four such planes. – 2017-01-19
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    Alternative generalizations ... Let a "trisohedralizer" of tetrahedron $OABC$ w.r.t. $O$ be a plane that meets edges $OA$, $OB$, $OC$ in $A^\prime$, $B^\prime$, $C^\prime$ such that tet $OA^\prime B^\prime C^\prime$ is "trisohedral": the three faces meeting at $O$ have equal area; we'll call the fourth face of that tetrahedron the *base* of the trisohedralizer. Then (barring degeneracies) there exists a unique "equal-area trisohedralizers point", where four trisohedralizers (one for each vertex) with a common base area concur. Likewise, there's a unique "equal-area parallelians point". – 2017-01-22

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