问题描述: 如图所示,AC平分∠DAB,AB>AD,CB=CD,CE⊥AB于E, (1)求证:AB=AD+2EB;(2)若AD=9,AB=21,BC=10,求AC的长. 1个回答 分类:数学 2014-10-03 问题解答: 我来补答 (1)证明:延长线段AD,过C作CF⊥AD交AD得延长线于F,∵AC为∠DAE的平分线,CE⊥AB,CF⊥AF,∴CE=CF,在Rt△CFD和Rt△CEB中CF=CECD=CB,∴Rt△CFD≌Rt△CEB(HL),∴FD=EB,又在Rt△CFA和Rt△CEA中CF=CEAC=AC,∴Rt△CFA≌Rt△CEA(HL),∴AF=AE,则AB=AE+EB=AF+EB=AD+DF+EB=AD+2EB;(2)∵AD=9,AB=21,由(1)得AB=AD+2EB,代入得9+2EB=21,解得EB=6,∴AE=AB-EB=21-6=15,又∵BC=10,在Rt△CEB中,根据勾股定理得:CE=BC2−BE2=8,在Rt△ACE中,根据勾股定理得:AC=AE2+CE2=17. 展开全文阅读