证明:AB=AC,则:∠B=∠C.(等边对等角);同理:AD=AE,则∠ADE=∠AED.∵∠ADC=∠B+∠BAD;即:∠ADE+∠CDE=∠B+∠BAD.∴∠AED+∠CDE=∠C+∠BAD.∴(∠CDE+∠C)+∠CDE=∠C+∠BAD.则:2∠CDE+∠C=∠C+∠BAD.故:∠BAD=2∠CDE.
