∵正五边形ABCDE中，已知△ABC面积为1，∴S△BCD=S△CDE=S△DEA=S△EAB=S△ACB=S△ACF=1．设S△AEF=x，则S△DEF=1-x，∵△AEF的边AF与△DEF的边DF上的高相等，∴DFAF＝1−xx．∵△DEF∽△ACF，∴S△DEFS△ACF＝(DFAF...