Let $N$ be the midpoint of the arc $BC$. Since $NI^2=NM.NK$, we have $NMI\sim NIK$ thus, $\angle KIN=\angle NMI$. \[\angle IKA=90-\angle AIK=90-(180-\angle KIN)=\angle NMI-90=\angle M'MI=\angle IM'M=180-\angle LM'I\]As desired.$\blacksquare$

Very similar (at least in terms of to what sort of points are similarities applied) to a problem from Bulgaria JBMO TST 2023.

Let $D$ be the antipodal point of $K$. By incenter lemma and shooting lemma, we have $$ID^2=CD^2=KD\cdot MD,$$ implying $\angle IKM=\angle DIM$. Since $\angle KAD=90^\circ=\angle BMD$ and $A,I,D$ are collinear, we have $$\angle IM’B=\angle IMB=\angle AKI,$$ so we are done.

Attachments:

Easy for P4…(will edit tmr) Set $(ABC)=\omega $ as the unit circle with $a=x^2$,$b=y^2$ and $c=z^2$ for not confusing with the imaginary unit $i$ let $j$ be the coordinate of $I$ in thr complex plane.$\boxed{j=-(xy+yz+zx)}$ and $2m=y^2+z^2$ if you draw the feet altitude from $I$ to $BC$ and reflect $M$ around that altitude you get $M’$ so doing calculations::: $\boxed{m’=\frac{y^2+z^2}{2}-xy-zx+\frac{yz(y+z)}{x}}$ since $K$ is the reflection of the point where angle bis. of $<BAC$ intersects $\omega $ with respect to the circumcenter and $O=0$ we get that $\boxed{k=yz}$.Since $A,K,B$ and $C$ all lie on $\omega $ We get that $AK\cap BC=L$ then $\boxed{l=\frac{x^2y^2+x^2z^2-x^2yz-y^2z^2}{x^2-yz}}$.Since you know all the points you calculate and get that $f(KLIM’)=f(\overline{KLIM’})$ $\boxed{\lambda }$

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