Did you mean $\angle EHL = \angle ABK + 30^\circ$? Let $O$ be the circumcenter of $\Delta ABC$ and $A_1$ be the foot of the remaining altitude. Then $L$ is the center of the circle containing points $B$, $H$, $O$, $C$ in this order. Since $HA_1 \parallel OL \perp BC$, by trivial angle chase this implies $\angle A_1HL = \angle HLO = 2\angle OCC_1 = 120^\circ - 2\angle C$. Now, $AE$ is the polar of $H$ with respect to the circumcircle $\omega$ of $CB_1C_1B$ centered at $M$, hence $H$ is the orthocenter of $\Delta MAE$ and $\angle EHA_1 = 90^\circ - \angle MAA_1 = \angle AMB = \angle C + \angle CAK$. Finally, we get $\angle EHL = \angle EHA_1 + \angle A_1HL = \angle CAK - \angle C + 120^\circ$. On the other hand, $\angle B_1BK = \angle B_1BC + \angle CBK = 90^\circ - \angle C + \angle CAK$, which completes the proof.