如图所示,AD是△ABC的角平分线,AD垂直平分线交AB于点E,交AC于点F,求证：四边形AEDF是菱形,速度,

连接DE、DF,∵EF垂直平分AD,∴AE=DE,AF=DF,
设AD与EF相交于G,在RTΔAGE与RTΔAGF中,AG=AG,∠EAG=∠FAG,∠AGE=∠AGF=90°
∴ΔAGE≌ΔAGF,∴AE=AF,∴AE=AF=DE=DF,
∴四边形AEDF是菱形.