I asked this question on MSE [here][1].


-----

Given a non-regular pentagon $A_1B_1C_1D_1E_1$, from the pentagon  $A_nB_nC_nD_nE_n$ construct the pentagon  $A_{n+1}B_{n+1}C_{n+1}D_{n+1}E_{n+1}$ where $A_{n+1}$ is the intersection between the angle bisector of $\angle D_n $ and $\angle C_n$, $B_{n+1}$ is the intersection between the angle bisector of $\angle D_n$ and $\angle E_n$, $C_{n+1}$ is the intersection between the angle bisector of $\angle A_n$ and $\angle E_n$, $D_{n+1}$ is the intersection between the angle bisector of $\angle A_n$ and $\angle B_n$ and $E_{n+1}$ is the intersection between the angle bisector of $\angle B_n$ and $\angle C_n$ **(the two opposite angles)**. I am allowing self-intersecting polygons in this constructions.  

My question is if this process is repeated indefinitely would the sequences ${A_n}$
, ${B_n}$
, ${C_n}$ , $D_n $ and $E_n$
 converge ?

There are only four possible scenarios:

 

 1. The points will converge.
 2. The points will eventually stuck on a loop.
 3. The points will completely diverge.
 4. Two or more points will eventually be coincident  which will end the sequence.




 

 
I tried to draw the first few pentagons to  see if the point will converge or not.

[![enter image description here][2]][2]
[![enter image description here][3]][3]

Here is the first 50 pentagons: 
[![enter image description here][4]][4]

I conjecture that the sequences will converge to a single point for all pentagons.  If this sequence converge then how to determine its limit given the initial pentagon?
 


  [1]: https://math.stackexchange.com/questions/4937560/does-the-sequence-formed-by-intersecting-angle-bisector-in-a-pentagon-converges
  [2]: https://i.sstatic.net/8M3RUVFT.png
  [3]: https://i.sstatic.net/2fbkfWiM.png
  [4]: https://i.sstatic.net/Ddnk9Px4.gif