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$I$ is the in-centre of $\triangle ABC$, and, $M,N$ the mid-points of the sides $AB,\, AC $.
$ BI \cap MN = K; \, CI \cap MN= L ; \, AL \cap BI= L_0 ; \, AK \cap CI = K_0$.
Prove that, the circumcircles of $\triangle ABL_0$ and $\triangle ACK_0$ touch each other.

Hint:

$\angle MKB=\angle KBC= \frac{1}{2}\angle B=\angle MBK $ , so $MB=MK$ . As in $\Delta AKB$ , $MA=MK=MB$ , $M$ is the center of the circumcircle of $\Delta AKB$ . M is the midpoint of AB , so it is a right-angle triangle - $\angle AKB=90^o$ . Similarly , $\angle ALC=90^o$ .
$\angle AKI+\angle ALI=180^o$ . hence quad $AKIL$ is cyclic .
Now , $\angle IAK = \angle ILK=\angle LCB = \frac{1}{2}\angle C=\angle K_0CA$ ,
$\angle IAL = \angle IKL=\angle KBC = \frac{1}{2}\angle B=\angle L_0BA$
by alternate segment theorem , $IA$ is tangent to the circumcircles of $\Delta AK_0C , \Delta AL_0B$ . As both have a common tangent to same point , they are tangent to each other .

Nice solution
It was inspired from JBMO 1997 #3.