如图,在正方形ABCD中,E是BC的中点,F是CD上一点,AE⊥EF,则S△ABE:S△ECF等于( )A. 1:2 B. 4:1 C. 2:1 D. 1:4 答案:【答案】B【解析】首先根据正方形的性质与同角的余角相等证得:△BAE∽△CEF,再根据相似三角形的性质可得结论.解:∵四边形ABCD是正方形,∴∠B=∠C=90°,AB=BC=CD,∵AE⊥EF,∴∠AEF=∠B=90°,∴∠BAE+∠AEB=90°,∠AEB+∠FEC=90°,∴∠BAE=∠CEF,∴△BAE∽△CEF,∴S△ABE:S△ECF=AB2:CE2,∵E是BC的中点,∴B() 5. After trying many times,they got to open the door . A. at the end B. at the end of C. in the end D. in the end of