在△ABC中,BD是角平分线,垂足为E,且AB=18cm,BC=12cm,S△ABC=90cm²,求DE的长,

过D做DF⊥AB于F
则DE=DF
S△ABC＝S△ABD+S△BDC＝(1/2)BC*DE+(1/2)AB*DF
=(1/2)*12*DE+(1/2)*18*DE=15DE=90
∴DE=6