Triangle angle bisectors, trisectors, quadrisectors, …

Question

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 ?

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          ^{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.}

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

Answer 1

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

Answer 2

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.

Answer 3

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.

Answer 4

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:

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Reference of above theorem:

(1) Euclid 4478.

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

(3) See Sriram Panchapakesan's message.