[asy][asy] size(6cm); pointpen=black; pair O=origin, A=dir(126.65), B=dir(-155.75), C=dir(-24.25), I=incenter(A, B, C), M=midpoint(B--C), N=midpoint(A--C), MA=midpoint(arc(O, C, B)), MB=midpoint(arc(O, A, C)), OA=midpoint(MB--N), W=midpoint(arc(OA, MB, N)); D(A--B--C--cycle, red); D(MA--MB, dotted); D(A--I--N, orange); D(unitcircle, deepblue); D(arc(OA, N, MB)--arc(OA, MB, N)--cycle, cyan+blue); markscalefactor=0.01; D(rightanglemark(N, I, A), black); D("A", A, A); D("B", B, B); D("C", C, C); D("I", I, dir(-60)); D("M", M, S); D("N", N, dir(60)); D("M_A", MA, MA); D("M_B", MB, MB); label("$\omega_A$", W, 2*W); [/asy][/asy] Let $\omega_A$ be the $A$-mixtilinear incircle of $\triangle ABC$, $T_A$ its tangency point with $\odot(ABC)$ and $M_A$ the midpoint of the arc $\overarc{BAC}$. It's well-known that $T_A, I, M_A$ are collinear. Then, $$\begin{aligned}AI\perp IN&\iff N\text{ is the tangency point of }\omega_A\text{ with }AC\\&\iff T_A=M_B\\&\iff M_A, I, M_B\text{ are collinear.}\end{aligned}$$Therefore, $$\begin{aligned}AI\perp IN&\iff BI\perp IM\\&\iff\angle CMI=\angle CIA\\&\iff AI\text{ is tangent to }\odot(IMC)\text{. }\blacksquare\end{aligned}$$

Useful one:https://artofproblemsolving.com/community/q1h410953p2306676

Here is a video in Chinese video

Let $X= IN\cap QC$. Let $F =(ABC)\cap AI$. Let $Q$ be the reflection of $F$ over $O$, the centre of $(ABC)$. We have \begin{align*} \measuredangle INM&= \measuredangle INC- \measuredangle MNC\\ &= \measuredangle QAC- \measuredangle BAC \\ &= \measuredangle QBC- \measuredangle BQC\\ &= \measuredangle QCB\\ &= \measuredangle XCM\implies MNXC\text{ is cyclic}. \end{align*} Also, \begin{align*} \measuredangle CMX&= \measuredangle CNX\\ &= \measuredangle CAQ \\ &= \measuredangle CBQ\\ &= \measuredangle QCB\\ &= \measuredangle XCM\implies MX=CX \implies X\text{ is the centre of } (CMQ). \end{align*} Now, since $$\measuredangle FIX=\measuredangle AIX=90^\circ=\measuredangle QCF=\measuredangle XCF$$and since $FI=FC$ and triangles $XIF$ and $XCF$ share a side, we conclude that $XI=XC$. Therefore, $I$ lies on $(CMQ)$ and now $FI^2=FC^2=FM\cdot FQ$ gives us that in fact $AI$ is tangent to $(IMC)$.

Let $AI\cap BC=D, \angle CAI=\alpha, \angle ICA=\beta, \angle CIM=\theta \implies \angle NIC=90-\alpha-\beta=\angle IBC$ and $\angle BIM=90+\alpha-\theta$. $IN$ and $IM$ are medians in $\triangle AIC$ an $\triangle BIC$. So from Since law in these triangles we get $$\frac{\sin(90)}{\sin(\alpha)}=\frac{\sin(90-\alpha-\beta)}{\sin(\beta)}=\frac{\sin(90+\alpha-\theta)}{\sin(\theta)} $$$$\implies \sin(90)\cdot \sin(\theta)=\sin(\alpha)\cdot \sin(90+\alpha-\theta) \implies \cos(90+\theta)=\cos(90+2\alpha-\theta) \implies \theta =\alpha$$$$\implies \angle DIM=\beta =\angle ICM $$$$\implies \text{AI is tangent to (IMC)}$$.