题目

如图24,已知ABCD是矩形纸片,E是AB上一点,BE∶EA =5∶3,EC=,把△BCE沿折痕EC翻折,若B点恰好落在AD边上,设这个点为F,图24(1)求AB、BC的长度各是多少;(2)若⊙O内切于以F、E、B、C为顶点的四边形,求⊙O的面积. 答案：思路分析:考察所给的条件,翻折△BCE,则△CBE≌△CFE,这样图形中提供了很多的线段相等、角相等.解:(1)连结CE、CF、EF,设BE =5x,EA =3x.∵四边形ABCD是矩形,∴AB =CD =8x,AD =BC,∠B =∠A =∠D =90°.∵△CBE≌△CFE,∴EF =5x,FC=BC,∠CFE =90°.∵∠AEF +∠EFC+∠DFC=180°,∴∠AFE +∠DFC=90°.又∵∠AEF +∠AFE =90°,∠AEF =∠DFC,∴sin∠AEF =sin∠DFC,即=.∴=,则FC =10x.∴ ==.∴x =3.∴AB =24,BC =30.(2)∵CE平分∠FCB和∠FEB,∴O在EC上.设⊙O和BC切于M,和AB切于N,连结OM、ON,设⊙O的半径为r,∴OM⊥BC,ON⊥AB.∴OM∥AB,ON∥BC.∴OM =BN =ON =BM =r.∴=,即=.∴r =10.∴⊙O的面积为100π.

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