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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.
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.
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