By angle chasing we have: $\angle IGC + \angle GCK = \angle DGC - \angle IGD + \angle GCB + \angle DCK = \angle GDB - \angle IDG + (180^\circ - 2\angle CDK) = \angle GDB - (90^\circ - \angle GDB) + (180^\circ - 2\angle CDK) = 90^\circ$ so that $GI \perp CL$ Let incircle $(I)$ touches $AC, AB$ at points $E, F$. We get $DEGF$ is a harmonic quadrilateral, yield tangent at $G$ of $(I), EF, BC$ are concurrent at a point name $S$ Obviously $(SD,BC)=-1$ then $-1=G(SD,BC)=G(SK,LC)$, on the other hand $GS \parallel CL$ therefore $K$ is the midpoint of $CL$

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Look at the figure above. Since $\angle CKG=\angle CKD=\angle CDK=\angle GDS=\angle SGD$ we get $SG\parallel CL$. Now $$-1=(B,C;S,D)\overset{\mathrm{G}}{=}(L,C;K,P_{\infty})$$So $K$ is the midpoint of $CL$