解：?$(1)$?如圖，直線?$DE、$?點?$F $?即為所求
?$(2)$?由題意可知，四邊形?$ABED$?與四邊形?$CFED$?關(guān)于直線?$DE$?對稱
∴?$∠DEB = ∠DEF$?
∵?$∠FEC = 48°，$?且?$∠DEB + ∠DEF = ∠FEC + 180°$?
∴?$2∠DEF = 48°+ 180°，$?解得?$∠DEF = 114°$?
∴?$∠DEC = ∠DEF - ∠FEC = 114° - 48° = 66°$?
?$(3)$?由題意可知，?$S_{\triangle BED}=S_{\triangle EDF}=4，$??$S_{\triangle AED}=S_{\triangle EDC}$?
設(shè)?$\triangle BED$?中邊?$BE$?上的高為?$h$?
則?$\frac {S_{\triangle BED}}{S_{\triangle EDC}}=\frac {\frac 12BE·h}{\frac 12EC·h}=\frac {BE}{EC}=\frac 25$?
∵?$S△{\triangle BED}=4，$?∴?$S_{\triangle EDC}=10$?
∴?$S△{\triangle AEC}=2S_{\triangle EDC}=20$?
設(shè)?$\triangle AEC$?中邊?$EC$?上的高為?$h'$?
∴?$\frac {S_{\triangle AEC}}{S_{\triangle ABC}}=\frac {\frac 12EC·h'}{\frac 12BC·h'}=\frac {EC}{BC}$?
∵?$\frac {BE}{EC}=\frac 25，$?則?$\frac {EC}{BC}=\frac 5{2 + 5}=\frac 57$?
∴?$S_{\triangle ABC}=\frac 75 S_{\triangle AEC}=\frac 75×20 = 28$?
