It's well-known (and can be prove by angle-chasing) that both $K,L$ lie on $DE$. Let the incircle $\omega$ of $\triangle{ABC}$ tangent to $AB$ at $F$. Let $IF$ intersect $\omega$ again at $T$. Note that $\angle{FDT}=90^{\circ}$. Combining with $BI\perp DF$ gives us $DT\parallel BI$. Hence, $\angle{ITD}=\angle{IDT}=\angle{DIB}$. Since $\angle{IDB}=\angle{IKB}=90^{\circ}$, we get that $IDKB$ concyclic. Therefore, $\angle{DIB}=\angle{HKD}$. Hence, $\angle{ITD}=\angle{HKD} \implies T\in (DKH)$. Similarly, we get that $T\in (ELH)$. This means $(DKH)$ and $(ELH)$ intersect at $T$, which lie on $\omega$, done.