1 ∠A=80° BD=BE CD=CF,求∠EDF
2. AB=AC AD =AE ∠BAD=30° 求∠EDC
(1)BD=BE,所以∠BDE=∠BED
∠B+2∠BDE=180
CD=CF,所以∠CDF=∠CFD
∠C+2∠CDF=180
两式相加:∠B+∠C+2(∠BDE+∠CDF)=360
两边除以2:∠BDE+∠CDF+(∠B+∠C)/2=180
因为∠BDE+∠CDF+∠EDF=180
所以∠EDF=(∠B+∠C)/2
∠A=80,所以∠B+∠C=100
因此∠EDF=50
(2)AB=AC,所以∠B=∠C
AD=AE,所以∠ADE=∠AED
∠AED为△CDE外角,所以∠AED=∠C+∠EDC
∠ADC为△ABD外角,所以ADC=∠B+∠BAD
因为∠ADC=∠ADE+∠EDC,所以∠C+∠EDC+∠EDC=∠B+∠BAD
2∠EDC=∠BAD=30
∠EDC=15