How can I prove that two lines intersect at a circle?---an extended observation. I’m referring to this problem:
How can I prove that two lines intersect at a circle??
I decided to graph it on GeoGebra. By accident, I did not draw it correctly. The line $RS$ was not the angle bisector, but still I had the intersecting point $I$ on the circle. Out of curiosity, I decided to rotate the line $RS$ and keep other relative relations, the intersecting point $I$ was always on the circle. I do not know how to solve it. Any suggestion?
Here I state the problem in the complete form:
A triangle $ABC$. Its inscribed circle touches the sides $AB, BC, CA$ in points $N, K$ and $M$ respectively. The lines $MN$ and $MK$ intersect any fixed line through $B$ in points $R$ and $S$. $RK$ and $SN$ intersect at the inscribed circle. When the line $RS$ rotates a complete round, the intersecting point $I$ will move around the inscribed circle completely one round.
 A: Now, I have the following attempt to my own question. The idea is from Michael Rozenberg in the original post.

In the picture, the exterior angle $B$ is equal to $2\angle SMR$. So $\angle RBN+\angle SBK=2\angle SMR$. Without loss of generality, we assume that $\angle RBN\leq\angle SBK$.
Draw the circle through $B,N,K$, intersecting $RB$ at $B'$. This circle also passes through $D$ because $D$ is the incenter. $BD$ is the diameter of this circle. $BD$ is the bisector of the angle $\angle ABC$. Hence arc $DN$=arc $DK$. So $\angle DB'B=90^\circ$, $\angle NB'D=\angle KB'D$. We can get $\angle RB'N=\angle SB'K=\angle SMR$.
Hence $R,B',K,M$ lie on a circle.  So $\angle B'RK=\angle KMB'$.
Similarly, we can have $S,B',N,M$ lie on a circle and $\angle NSB'=\angle NMB'$.
The exterior angle $\angle SIK=\angle B'SI+\angle B'RI=\angle SMR$, so four points $M,N,I,K$ lie on a circle.
It is almost the complete steps of Michael Rozenberg's answer. I just moved $B$ to $B'$.
So far, I do not see how to show that the trace of $I$ is the complete inscribed circle.
