22.(期中·河北区)(本小题10分)如图,四边形ABCD中,AB=AC,∠D=90°,BE⊥AC于点F,交CD于点E,连接EA,EA平分∠DEF.
(1)求证:AF=AD.
(2)若BF=7,DE=3.求EF和CE的长.

22. (1)【证明】
∵ BE⊥AC,
∴ ∠AFE=90°=∠D.
∵ EA平分∠DEF,
∴ ∠AEF=∠AED.
∵ AE=AE,
∴ △AFE≌△ADE(AAS),
∴ AF=AD. (2)【解】由(1)知△AFE≌△ADE,
∴ EF=DE=3.
∵ BE⊥AC,
∴ ∠BFA=90°=∠D.
∵ AD=AF,AB=AC,
∴ Rt△AFB≌Rt△ADC(HL),
∴ CD=BF=7,
∴ CE=CD−DE=4.