For starters, we know that a trirectangular tetrahedron is a tetrahedron where all three face angles at one vertex are right angles.
Here's an interesting property I stumbled upon, of which I look for an elegant proof:
(For the sake of explanation, I shall model this tetrahedron in the Cartesian plane, with origin as a vertex and the three normal vectors of the three faces intersecting at the origin, along the coordinate axes)
And here it goes
Consider the plane, $$ \frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1 $$ forming a tetrahedron OABC, with the XY, YZ, and ZX planes. O is the origin, and A,B,C are the other three vertices. The property is, that, the foot of perpendicular from O, on plane ABC, is the orthocentre of triangle ABC.
I have already proved it, with an analytic approach (extending the Cartesian model I describe above), however, I'm looking for a rather elegant, and possible geometric proof for this property.
Please share your ideas and proofs, thanks a lot!
P.S. Feel free to ask for the analytic proof (It's probably too obvious, and also undesirably lengthy) which I've completed. (I shall attach a picture of the handwritten work to this post)
Let $OABC$ be your tetrahedron. $OA$ is perpendiclular to $OB$ and $OC$, so $OA$ is perpendicular to the plane $OBC$ and, in particular, $OA \perp BC$. Now, the theorem of three perpendiculars states that if some line is perpendicular to $l$, then its projection onto any plane containing $l$ is also perpendicular to $l$. So, if $H$ is the foot of perpendicular from $O$ onto $ABC$, then $AH\perp BC$.
By the way, tetrahedrons such that each pair of opposite edges are perpendicular are called orthocentric tetrahedrons. Your fact is true for any orthocentric tetrahedron.