TITLE: Tetrahedron angles sum to $\pi$: Bisector plane
QUESTION [5 upvotes]: I discovered empirically what to me is an amazing lemma
concerning face angles of a tetrahedron.
Let $\triangle abc$ be a triangle in the $xy$-plane, and $d$ the
apex of a tetrahedron with positive $z$ coordinate.
The lemma is this:

Lemma. The locus of points $d$ for which the sum
  of the nonbase tetrahedron face angles incident to $b$ sum to $\pi$,
  $$\angle dba + \angle dbc = \pi$$
  is a vertical (parallel to $z$) bisector plane which meets the $xy$-plane
  in a line $L$ that has the property
  that the angle of incidence $\beta$ between $ab$ and $L$ is equal
  to the angle of reflection $\beta$ between $bc$ and $L$.


 
 
 
 
 


If I express this relationship in terms of ArcCos( )'s of the relevant
angles, it all works out algebraically/trigonometrically.  So I have a "proof" in this
(limited) sense.
But surely for such a simple angle of incidence = angle of reflection
relationship there is a concise geometrical explanation—maybe involving reflecting light rays...?
Amidst a much longer proof, this was at one point a critical lemma, but now I have circumvented
its need (I think?!).  Nevertheless, it would be illuminating to see a more revealing proof.  Thanks for ideas and/or insights!

REPLY [3 votes]: Reflect $a$ across $L$ in the $xy$-plane to $a'$, so $\angle a'bc$ is a straight angle, and hence $\angle dba' + \angle dba = \pi$. If $d$ is in the bisector plane then $\angle dba' = \angle dba$, if $d$ is on the same side of the bisector plane as $a$ then $\angle dba' > \angle dba$, and if $d$ is on the opposite side then $\angle dba' < \angle dba$.