AB=AC ,则∠C=∠B=65°
所以∠BAC=180°-2X65°=50°
∠DAC=∠BAC-∠BAD=50°-20°=30°
AD=AE,则∠ADE=∠AED=(180°-30°)÷2=75°
∠AED=∠C+∠EDC
∠EDC=∠AED-∠C=75°-65°=10°
2.∠DBA=20°,∠EDC=∠CDE=10°
∠DBA=2∠CDE
AB=AC ,则∠C=∠B=65°
所以∠BAC=180°-2X65°=50°
∠DAC=∠BAC-∠BAD=50°-20°=30°
AD=AE,则∠ADE=∠AED=(180°-30°)÷2=75°
∠AED=∠C+∠EDC
∠EDC=∠AED-∠C=75°-65°=10°
2.∠DBA=20°,∠EDC=∠CDE=10°
∠DBA=2∠CDE