$AD\cap EK=S$ $\angle BDS=90=\angle BKS\implies B,D,K,S$ are cyclic. We have $EK\perp BH\perp AC\implies EK\parallel AC$ $\angle ABK=\angle CAK=\angle EKA$ and $\angle KAS=\angle KAL=\angle KBL$ Hence $180-\angle LST=\angle KSL=\angle KAS+\angle SKA=\angle KBL+\angle ABK=\angle ABL$ $\implies T,S,L,B$ are cyclic. $\angle BDK=\angle BST=\angle BLT$ a desired.$\blacksquare$

Claim: Triangles $BDL$ and $BKT$ are similar $$\angle BDL=\angle BKT=90^{\circ}$$ $$\angle LBD=\angle ALB-90^{\circ}=90^{\circ}-\angle AEB=90^{\circ}-\angle BAC=\angle ABH=\angle TBK$$ Then by SST triangles $BDK$ and $BLE$ are also similar, proving the equality.

Attachments:

Prove ET parallel to AC and then prove triangle TBL is similar to triangle KBD

it is easy to see that $\angle DBH = \angle CAH = \angle LBT$ so $\angle DBL = \angle KBT$ and $\angle BDL = \angle BKT$ then $\triangle BDL\sim\triangle BKT$ then by the spiral similarity lemma $\triangle BDK\sim\triangle BLT$ so $\angle BDK = \angle BLT$