$ (1)$解：如图，连接$BG$
∵$∠EAD = 90°，$$∠BAC = 30°$
∴$∠DAG=∠EAD+∠BAC = 120°$
∵$∠ADE = 30°，$$∠ADE+∠DAG+∠AG D = 180°$
∴$∠AG D = 180°-∠ADE-∠DAG = 30°，$即$∠AG D=∠ADE = 30°$
∴$AG = AD$
∵$AB = AD，$∴$AG = AB，$即$∠ABG=∠AG B$
∵$∠BAC+∠ABG+∠AG B = 180°$
∴$∠AG B=∠ABG=\frac 12(180°-∠BAC)=75°$
∵$BH\perp DF，$∴$∠FHB=∠BHE = 90°，$即$∠BHE=∠EAD$
$ $又$∠BEH=∠AED，$$∠BHE+∠BEH+∠EBH = 180°，$
$∠EAD+∠AED+∠ADE = 180°，$∴$∠EBH=∠ADE = 30°$
∴$∠HBG=∠ABG-∠EBH = 45°$
∴$∠HG B = 90°-∠HBG = 45°，$即$∠HG B=∠HBG$
∴$GH = BH$
$ $又$AH = AH，$∴$\triangle AGH≌\triangle ABH(\mathrm {SSS})$
∴$∠HAG=∠HAB$
$ $又$∠HAG+∠HAB=∠BAC，$∴$∠HAE = 15°$
$ (2)$证明：如图，在$DH$上取点$M，$使$HM = HF，$连接$BM$
∵$∠ABC=∠EAD = 90°，$∴$AD// BF$
∴$∠F=∠ADE = 30°$
∵$BH\perp DF，$$HM = HF，$∴$F B = BM$
∴$∠BMF=∠F = 30°$
∵$AB = AD，$$∠EAD = 90°$
∴$\triangle BAD$是等腰直角三角形，即$∠ADB = 45°$
∴$∠MDB=∠ADB-∠ADE = 15°$
∵$∠BMF=∠MBD+∠MDB$
∴$∠MBD=∠BMF-∠MDB = 15°，$即$∠MBD=∠MDB$
∴$BM = DM，$即$DM = F B$
∵$DH = DM+HM，$∴$DH = F B+FH$
