如图,等腰梯形ABCD中,AD=2,BC=6,高DF=2,则腰长DC=________.
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解析分析:过A作AE⊥BC于E,证平行四边形ADFE和△AEB≌△DFC,推出EF=AD,AE=DF,求出CF长,根据勾股定理即可求出CD.
解答:解:过A作AE⊥BC于E,∵DF⊥BC,∴∠AEB=∠DFC=90°,DF∥AE,∵AD∥BC,∴四边形ADFE是平行四边形,∴AD=EF=2,AE=DF∵AD∥BC,AB=CD,∴∠B=∠C,∵AE=DF,∠AEB=∠DFC,∴△AEB≌△DFC,∴BE=CF=(BC-AD)=2,在△DFC中,由勾股定理得:DC==2,故