24.如圖1,在等腰Rt△ABC中,∠ABC=90°,AC=9.點(diǎn)D為AC邊上的點(diǎn),連結(jié)BD,在邊BC下方作△BCE,使得△BCE≌△ABD,且∠BAD=∠BCE,在射線EC上取一點(diǎn)F,使得DF=EF,連結(jié)DE.
(1)求證:∠BEF=∠BDF.
(2)當(dāng)AD=3時(shí),求CF的長(zhǎng)度.
(3)如圖2,在AC右側(cè)作GA⊥AC于點(diǎn)A,延長(zhǎng)BD交AG于點(diǎn)H,當(dāng)△DFH是以HF為腰的等腰三角形時(shí),則DF的長(zhǎng)為
(直接寫(xiě)出答案).