解：?$(2)$?設(shè)?$∠BAG = x。$?
∵?$AE\perp AB，$?
∴?$∠EAG = 90°-∠BAG = 90°-x。$?
∵?$AG $?平分?$∠EAM，$?
∴?$∠EAM = 2∠EAG = 180°-2x，$?
∴?$∠BAM = 90°-∠EAM = 2x - 90°。$?
∵?$MN// PQ，$??$AB// CD，$?
∴?$∠ABQ=∠BAM，$??$∠CDQ=∠ABQ，$?
∴?$∠CDQ=∠BAM = 2x - 90°。$?
∵?$CD\perp DF，$?
∴?$∠FDQ = 90°+∠CDQ = 2x，$?
∴?$2∠BAG=∠FDQ。$?
?$ (3)∠HBD+∠AHB+∠BAH = 240°$?或
?$∠AHB+∠BAH-∠HBD = 120°。$?
理由如下：
如圖①，當(dāng)點?$H$?在點?$K$?上方時，過點?$H$?
作?$HT// MN，$?則?$HT// MN// PQ，$?
∴?$∠1=∠HBD，$??$∠MAB=∠ABD = 60°，$?
?$∠AHT+∠HAM = 180°，$?
∴?$∠HBD+∠AHB+∠HAM = 180°，$?
∴?$∠HBD+∠AHB+∠HAM+∠MAB = 240°，$?
即?$∠HBD+∠AHB+∠BAH = 240°；$?
如圖②，當(dāng)點?$H$?在點?$C，$??$K$?之間時，過點?$H$?作
?$HT// MN，$?則?$HT// MN// PQ，$?
∴?$∠HBD=∠THB，$??$∠THA=∠HAC，$?
?$∠BAC = 180°-∠ABD = 120°，$?
∴?$∠HBD=∠THA+∠AHB=∠AHB+∠HAC$?
∴?$∠HBD=∠AHB+∠BAH-∠BAC，$?
∴?$∠AHB+∠BAH-∠HBD=∠BAC，$?
即?$∠AHB+∠BAH-∠HBD = 120°；$?
如圖③，當(dāng)點?$H$?在點?$C，$??$D$?之間時，過點?$H$?作
?$HT// MN，$?則?$HT// MN// PQ，$?
∴?$∠HAN=∠AHT，$??$∠BHT=∠HBD，$?
?$∠BAC = 180°-∠ABD = 120°，$?
∴?$∠AHT = 120°-∠BAH，$?
∴?$∠AHB=∠AHT+∠BHT $?
?$= 120°-∠BAH+∠HBD，$?
∴?$∠AHB+∠BAH-∠HBD = 120°。$?
當(dāng)點?$H$?在點?$K$?或點?$C$?處時，經(jīng)檢驗，符合
?$∠AHB+∠BAH-∠HBD = 120°。$?
綜上所述，滿足條件的關(guān)系是
?$∠HBD+∠AHB+∠BAH = 240°$?或
?$∠AHB+∠BAH-∠HBD = 120°。$?
