三角形中∠ABC=∠ACB,D为BC上一点,E为直线AC上一点,且∠ADE=∠AED;求证:∠BAD=2∠CDE

∠CDE+∠ACB=∠AED(因为∠ABC=∠ACB)所以∠CDE+∠ABC=∠AED∠ABC+∠BAD=∠ADC=∠ADE+∠EDC(因为∠ADE=∠AED)∠ABC+∠BAD=∠AED+∠EDC=∠EDC+∠ABC+∠EDC即∠ABC+∠BAD=∠CDE+∠ABC+∠CDE消去∠ABC等∠BAD=∠CDE+∠CD...