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 $A_{n+1}B_{n+1}C_{n+1}D_{n+1}E_{n+1}$ as follows: - $A_{n+1}$ is the intersection between the angle bisector of $\angle C_n $ and $\angle D_n$. - $B_{n+1}$ 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 $\angle E_n$ and $\angle A_n$. - $D_{n+1}$ is the intersection between the angle bisector of $\angle A_n$ and $\angle B_n$. - $E_{n+1}$ is the intersection between the angle bisector of $\angle B_n$ and $\angle C_n$. **(the two opposite angles)** The reason that I chose this construction is that the point $A_{n+1}$ is the only point that doesn't depend on the position of $A_n$. 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 trapped on a loop. 3. The points will diverge completely. 4. Two or more points coincide, ending 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 in a zoom-in animation: [![enter image description here][4]][4] I conjecture that the sequences converge to a single point for all non-regular pentagons. If convergence occurs, how can we determine the limit point based on the initial pentagon? --- [Here][5] is a Geogebra file that have the first 50 pentagons. [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 [5]: https://www.geogebra.org/classic/tywbvnyg