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如图,在△ABC中,∠C=2∠B,AD是△ABC的角平分线。求证:AB=AC+CD
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解:延长AC到E,使CE=CD,连接DE,
∴∠CDE=∠CED,
∵∠ACB=∠CDE+∠CED,
∵∠ACB=∠CDE+∠CED,
∴∠ACB=2∠CED,
∵∠C=2∠B,
∴∠B=∠E,
∵AD为△ABC的角平分线,∴∠BAD=∠CAD,
∵AD=AD,
∴△ABD≌△AED,
∴AB=AE,
∴AB=AC+CD.
∴∠CDE=∠CED,
∵∠ACB=∠CDE+∠CED,
∵∠ACB=∠CDE+∠CED,
∴∠ACB=2∠CED,
∵∠C=2∠B,
∴∠B=∠E,
∵AD为△ABC的角平分线,∴∠BAD=∠CAD,
∵AD=AD,
∴△ABD≌△AED,
∴AB=AE,
∴AB=AC+CD.
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