證明:作∠HCD=10°,交DE於G,交BE於F,連線DF∵AB=AC,∴∠ABC=∠ACB,∵∠A=20°,∴∠ABC=∠ACB=(180°-20°)/2=80°,∵∠BCD=50°∵∠HCD=10°∴∠HCB=60°∵∠FBC=60°∴△BCF是等邊三角形∴BC=BF∵∠BCD=50°∵∠DBC=80°∵∠DBC+∠BCD+∠BDC=180°∴∠BDC=50°∵∠BCD=50°∴∠BDC=∠BCD∴BD=BC∴BD=BF∴∠BDF=∠BFD∵∠DBF=80°－∠FBC(60°)=20°∴∠BDF=80°∵∠BDC=50°∴∠CDF=30°∴∠DFH=∠CDF(30°)+∠FCD(10°)=40°∵∠DHF+∠DFH(40°)=∠BDF(80°)∴∠DHF=40°∵∠DFH=40°∴∠DHF=∠DFH∴DH=DF∵BC=BC∵∠ABC=∠ACB∵∠HCB=∠EBC∴△HBC≌△ECB∴HC=EB∵BF=CF∴HF=EF∵∠HFE=∠BFC=60°∴△HFE是等邊三角形∴HE=FE∵DH=DF(已證)∵DE=DE∴△DHE≌△DFE∴∠HDE=∠FDE∵∠DHF(40°)+∠FHE(60°)+∠HEF(60°)+∠EFH(60°)+∠HFD(40°)+∠HDE+∠FDE=360°∴∠EDF=50°∵∠CDF=30°∴∠EDC=80°∴∠DEB=50°+60°-80°=30°