(1)
因为△CAD≌△CED,所以∠1 = ∠2.
因为△CEF≌△CAD,所以∠2 = ∠3.
而∠ACB = 90°,所以∠1 = ∠2 = ∠3 = 90°÷3 = 30°.
因为△CEF≌△BEF,所以∠B = ∠3 = 30°.
(2)
因为△CEF≌△BEF,所以∠CFE = ∠BFE.
而∠CFE + ∠BFE = ∠CFB = 180°,所以∠BFE = 90°.
又∠BCA = 90°,所以EF∥AC.
(1)
因为△CAD≌△CED,所以∠1 = ∠2.
因为△CEF≌△CAD,所以∠2 = ∠3.
而∠ACB = 90°,所以∠1 = ∠2 = ∠3 = 90°÷3 = 30°.
因为△CEF≌△BEF,所以∠B = ∠3 = 30°.
(2)
因为△CEF≌△BEF,所以∠CFE = ∠BFE.
而∠CFE + ∠BFE = ∠CFB = 180°,所以∠BFE = 90°.
又∠BCA = 90°,所以EF∥AC.