I asked this question on MSE here.

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.

Here is the first 50 pentagons:

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?