Triangle angle bisectors, trisectors, quadrisectors, … With the triangle
angle bisector theorem
and
Morley's trisector theorem
as background ,
are there any pretty theorems known for triangle $n$-sectors,
$n > 3$?
For example, angle quadrisectors?
The  images  below  suggest a THEOREM which I'm hesitant to believe,
but illustrates what I seek :

Q1. Do the $\tfrac{1}{4}$ rays (brown) meet the $\tfrac{1}{2}$ rays (red) as suggested
by the black segments, or is that only approximately true ?


          


         

Quadrisectors. Center: Equilateral triangle. Left & Right: Altitude fixed.


         

Base length: $1$. Altitude: $\sqrt{3}/2$.


         


         

Left figure enlarged, showing apparent coincidence between
half and quarter angle rays.



Q2. Are there any "nice" theorems known for how rays $n$-sectoring
the angles of a triangle meet one another?

 A: Q1:
The "low" angle quadrisectors coming from $B$ and $C$ (i.e. the ones closer to $\overline{BC}$) meet on the angle bisector coming from $A$ iff $ABC$ is isosceles (with $AB = AC$). 
Proof:
The difficult proof is the "only if". Let $O$ be the incenter of $ABC$, where the bisectors meet. Then the angle quadrisectors of $ABC$ are the angle bisectors of $OBC$ - so they meet at a point on the angle bisector coming from $O$. Therefore, if they meet on the angle bisector coming from $A$, then $\overline{AO}$ bisects the angle $\angle BOC$. But then by supplementary angles, $\angle AOB \cong \angle AOC$ - and by the definition of angle bisector, $\angle OAB \cong \angle OAC$, and $\overline{OA} = \overline{OA}$, so by ASA, $\triangle AOB \cong \triangle AOC$, and $\overline{AB} \cong \overline{AC}$.
A: The Lighthouse Theorem of the late Richard Guy says two sets of $n$ lines at equal angular distances, one set through each of the points $B$, $C$, intersect in $n^2$ points that are the vertices of $n$ regular $n$-gons. The circumcircles of the $n$-gons each pass through $B$ and $C$.
Guy, Richard K. (2007), "The lighthouse theorem, Morley & Malfatti—a budget of paradoxes,"  American Mathematical Monthly, 114 (2): 97–141, JSTOR 27642143, MR 2290364.
A: Q2: If you take the lines from each vertex to the quadrisectors of their opposite sides, then concurrency does occur giving a hexagon in the middle which is always $\dfrac 8 {35}$ of the triangle area. This can easily be proved with coordinate geometry.
A: Q. 2:
Let $\triangle ABC$ be any triangle and let $O$, $H$ be the circumcenter and orthocenter of $\triangle ABC$ respectively.  Then the trisectors of $\angle OAH$, $\angle OBH$, $\angle OCH$ closest to the lines $HA$, $HB$, $HC$ bound an equilateral triangle as Shown in figure given below:



Reference of above theorem:
(1) Euclid 4478.
(2) Euclid messenger group discussion, related to above theorem.
(3) See Sriram Panchapakesan's message.
