题目

如图,在矩形ABCD中,E是AD边的中点,BE⊥AC,垂足为点F,连接DF,下列结论:①△AEF∽△CAB;②CF=2AF;③tan∠CAD=.其中正确的结论有 ( ) A. 3个 B. 2个 C. 1个 D. 0个 答案:【答案】B【解析】①正确.只要证明∠EAC=∠ACB,∠ABC=∠AFE=90°即可;②正确.由AD∥BC,推出△AEF∽△CBF,推出,由AE=AD=BC,推出=,即CF=2AF;④错误,设AE=a,AB=b,则AD=2a,由△BAE∽△ADC,有,即b=a,可得tan∠CAD==即可得.如图,过D作DM∥BE交AC于N,∵四边形ABCD是矩形,∴AD∥BC,∠ABC=90°,AD=BC,∵BE⊥AC于 In that school, English is compulsory for all students, but French and Russian are ______.A.specialB.regionalC. optionalD.original
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