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證明:∵△ABC是等邊三角形,BD是中線,∴∠ABC=∠ACB=60°.∠DBC=30°(等腰三角形三線合一).又∵CE=CD,∴∠CDE=∠CED.又∵∠BCD=∠CDE+∠CED,∴∠CDE=∠CED=12∠BCD=30°.∴∠DBC=∠DEC.∴DB=DE(等角對等邊).
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