Let $(I)$ touch $BC$ at $R.$ Since $\angle RIC+\angle LKI=\angle RIC+\angle ICR=90^{\circ},$ it follows that $RI$ passes through the circumcenter of $\triangle LIK.$ Therefore, if $\odot(ILK)$ is tangent to $(I)$ through $N,$ then $N$ is the antipode of $R$ WRT $(I).$ Let line $CN$ cut $(I)$ again at $T.$ Then we have $\angle PNT=\angle RNC+\angle RNP=\angle RNC+\angle NCR=90^{\circ}.$ Hence, $T$ is the antipode of $P$ WRT the incircle $(I)$ $\Longrightarrow$ $N$ is the Nagel point of $\triangle ABC.$ Now, it is well-known that $N$ lies on the small arc $PQ$ of $(I)$ if and only if $AB+AC=3BC.$ The inverse excercise is proved analogously. Condition $AB+AC=3BC$ implies that the antipode $N$ of $R$ WRT $(I)$ coincides with the Nagel point of $\triangle ABC.$ Then $BI \cap NC$ coincides with $K$ (orthogonal projection of C onto BI) $\Longrightarrow$ $N$ is the antipode of $I$ WRT the circle $\odot(ILK)$ $\Longrightarrow$ $\odot(ILK)$ is tangent to $(I)$ through $N.$

As I heard there are several results from condition $ |AB|+|AC|=3|BC| $ , Can someone help me with finding some pages about it ?

@mahanmath. These are the ones that I know, there must be additional results 1) $IG \perp BC,$ where $I,G$ are the incenter and centroid of $\triangle ABC.$ 2) Nagel point lies on the incircle $(I).$ Further, it's the antipode of $R \equiv (I) \cap BC.$ 3) Mittenpunkt $X_{9}$ of $\triangle ABC$ lies on the A-sideline of its incentral triangle. 4) Circumcenter $O$ is equidistant from $I$ and the midpoint of $\overline{IA}.$ We have some metric and trigonometric relations such as: $AI^2=\frac{ AB \cdot AC}{2}=4Rr \ , \ r= \frac{r_a}{2}=\frac{IB \cdot IC}{IA}$ $\cot \frac{A}{2}= \cot \frac{B}{2}+ \cot \frac{C}{2} \ , \ \tan \frac{B}{2} \cdot \tan \frac{C}{2}=\frac{1}{2}$ $\sin \frac{A}{2}=\sin \frac{B}{2} \cdot \sin \frac{C}{2}=\frac{1}{3} \cos \left (\frac{B-C}{2} \right)=\sqrt{\frac{r}{4R}}$

Thanks a lot !