qweDota wrote: Let $N,K,L$ be points on $AB,BC,CA$ such that $CN$ bisector of angle $\angle ACB$ and $AL=BK$.Let $BL\cap AK=P$.If $I,J$ be incenters of triangles $\triangle BPK$ and $\triangle ALP$ and $IJ\cap CN=Q$ prove that $IQ=JP$ https://artofproblemsolving.com/community/c6h1573838p9678185

For $BK$ and $AL$, the spiral similarity center should be considered.

Define $M = (ALP) \cap (BKP)$ as the center of spiral congruence mapping $\triangle AML \cong \triangle KMB$. Suppose $IJ \cap (ALP) = X$ and $IJ \cap (BKP) = Y$. Note $M$ is equidistant from $AL$ and $BK$, so it lies on the $C$ bisector. The congruences give $MX=MY$ as the corresponding $M$-bisectors. $XI = YJ$ through incenter-excenter lemma. Thus it suffices to show $MP=MQ$, or $\angle P = \angle Q$, which is a short angle chase. $\blacksquare$