Consider a tetrahedron with a right-angle corner (like the corner of a cube) at $D$. Prove that the foot of the height from $D$ to the face ABC is the orthocenter of $ABC$.
Is this well-known?
The foot of the height in an orthocentric tetrahedron is the orthocenter of a face.
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geometry
1 Answers
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$BD\perp DAC$ (plane) so in particular $BD\perp AC$
$DH\perp ABC$ (plane) so $DH\perp AC$
So $AC \perp BD$ and $AC\perp DH$ so $AC\perp DHB$ (plane). In particular $AC\perp BH$, so $BH$ is altitude. Similarly for the other vertices, so $H$ is orthocenter.
And yes, this is well-known (at least by any participant in a Math Olympiad)