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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 ?


         

Quad3Tri
          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
          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?

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    $\begingroup$ For Q1, it looks like your suggestion is that the "corresponding" quadrisectors meet at a point on the bisector, right? If so, then the left and right triangles seem to contradict the suggestion when looking at the non-black-lined bisector. $\endgroup$
    – user44191
    Nov 16, 2019 at 0:42
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    $\begingroup$ @user44191: Superficially, it seems moving the apex left of the equilateral position maintains one set of intersections, and moving right maintains the other set of intersections (both indicated by the endpoints of the black segment). $\endgroup$ Nov 16, 2019 at 2:23
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    $\begingroup$ But that can't be correct: the apparent coincidence is some algebraic identity, so if it fails when you move "right" in some direction then it can't hold exactly when you move "left" in the same direction. $\endgroup$ Nov 16, 2019 at 2:56
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    $\begingroup$ Your left and right triangles are too close to isosceles. Here is a GeoGebra drawing where you can drag the points around to see that for "most" triangles, the quadrisectors don't intersect on the bisector: geogebra.org/geometry/kgsh3ch5 $\endgroup$ Nov 16, 2019 at 3:49
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    $\begingroup$ Thanks @ZachTeitler and NoamElkies and user44191 for uncovering that the coincidence requires the triangle to be isosceles. $\endgroup$ Nov 16, 2019 at 12:08

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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 $\overline{AB} \cong \overline{AC}$.

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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.

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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.

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    $\begingroup$ I think this was downvoted because the question asked more about the quadrisectors of the angles than of the sides. $\endgroup$
    – Matt F.
    Mar 6 at 22:42
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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:


Angle Trisector makes Equilateral triangle


Reference of above theorem:

(1) Euclid 4478.

(2) Euclid messenger group discussion, related to above theorem.

(3) See Sriram Panchapakesan's message.

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    $\begingroup$ Please use Markdown *Markdown*, not $MathJax$ $MathJax$, for emphasis. I have edited accordingly. $\endgroup$
    – LSpice
    Mar 6 at 20:26

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