已知等边△ABC纸片,将它折叠,使点A落在BC边的点A1位置上,折痕为EF,点E、F分别在AB、AC上,若BA1:CF=2:3,求AE:AF的值图:file://C:\Documents and Settings\Administrator\Local Settings\Temporary Internet Files\Content.IE5

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已知等边△ABC纸片,将它折叠,使点A落在BC边的点A1位置上,折痕为EF,点E、F分别在AB、AC上,若BA1:CF=2:3,求AE:AF的值图:file://C:\Documents and Settings\Administrator\Local Settings\Temporary Internet Files\Content.IE5

已知等边△ABC纸片,将它折叠,使点A落在BC边的点A1位置上,折痕为EF,点E、F分别在AB、AC上,若BA1:CF=2:3,求AE:AF的值图:file://C:\Documents and Settings\Administrator\Local Settings\Temporary Internet Files\Content.IE5
已知等边△ABC纸片,将它折叠,使点A落在BC边的点A1位置上,折痕为EF,点E、F分别在AB、AC上,若BA1:CF=2:3,求AE:AF的值
图:file://C:\Documents and Settings\Administrator\Local Settings\Temporary Internet Files\Content.IE5\KNKF8K7Y\20100508211149-1263032507[1].jpg

已知等边△ABC纸片,将它折叠,使点A落在BC边的点A1位置上,折痕为EF,点E、F分别在AB、AC上,若BA1:CF=2:3,求AE:AF的值图:file://C:\Documents and Settings\Administrator\Local Settings\Temporary Internet Files\Content.IE5
由题意可知,△AEF ≌ △A1EF
∴ AE:AF = A1E:A1F ∠EA1F=60° ∠B=60° ①
对△BA1E,∠EA1C = ∠BEA1+∠B ②
又 ∠EA1C = ∠EA1F+∠FA1C ③
由①②③ 得 ∠BEA1= ∠FA1C ④
∠B = ∠C ⑤
、根据④⑤两个角相等的关系
△BA1E ∽ △CFA1
有相似三角形得到对应成比例关系
BA1:CF = EA1:A1F=2 :3
所以 AE:AF=EA1:A1F= 2 :3