Since $\angle CKE+\angle BKF=\angle CLE+\angle BLF=180^{\circ}$, have $$\angle ELF+\angle BLC=\angle EKL+\angle BKC=180^{\circ}.$$Since $$\angle ELF=\angle EKF=\angle EDF=\frac{\angle EIF}{2}=\frac{180^{\circ}-\angle BAC}{2}=90-\frac{\angle BAC}{2},$$thus $$\angle BLC=\angle BKC=90+\frac{\angle BAC}{2}=\angle BIC,$$therefore $BLIKC$ is cyclic. Now, let us apply inversion with centre $I$ and circle $\omega$, since $B\mapsto M_3,$ midpoint of $DF$, and $C\mapsto M_2,$ midpoint of $DE$. Thus, $$\angle IM_3K=\angle IKB=\angle ICB=\angle ICA=\angle ICE=\angle IDE=\angle IDM_2=\angle IM_3M_2,$$thus $M_2,M_3$ and $K$ are collinear, similarly $M_2,M_3$ and $L$ are collinear. Since trivially $D,E,F$ are equidistant from $M_2M_3$ (for example, one could use parallelity and homothety), we are done.

we show by angle chase that $K,L\in(BIC)$ so to prove the result it suffices to show that $ED,DF$ cuts $KL$ at theirs midpoints say $U,V$ ; but $U\in ED \cap BI$ and $BIDE$ is cyclic thus $UB.UI=UE.UD$ which leads $U$ is on the radical axis $KL$ similarly we get $V\in KL$. RH HAS