(1)∵AB=AC,EC=ED,∠BAC=∠CED=60°, ∴△ABC∽△EDC, ∴∠CBD=∠CAE, ∴∠AFB=180°-∠CAE-∠BAC-∠ABD =180°-∠BAC-∠ABC =∠ACB, ∴∠AFB=60°, 同理可得:∠AFB=45°;
(2)∵AB=AC,EC=ED,∠BAC=∠CED, ∴△ABC∽△EDC, ∴∠ACB=∠ECD,=, ∴∠BCD=∠ACE, ∴△BCD∽△ACE, ∴∠CBD=∠CAE, ∴∠AFB=180°-∠CAE-∠BAC-∠ABD, =180°-∠BAC-∠ABC=∠ACB, ∵AB=AC,∠BAC=α, ∴∠ACB=90°-α, ∴∠AFB=90°-α. 故答案为:∠AFB=90°-α.
(3)图4中:∠AFB=90°-α; 图5中:∠AFB=90°+α. ∠AFB=90°-α的证明如下: ∵AB=AC,EC=ED,∠BAC=∠CED, ∴△ABC∽△EDC, ∴∠ACB=∠ECD,=, ∴∠BCD=∠ACE, ∴△BCD∽△ACE, ∴∠CBD=∠CAE, ∴∠AFB=180°-∠CAE-∠BAC-∠ABD, =180°-∠BAC-∠ABC=∠ACB, ∵AB=AC,∠BAC=α, ∴∠ACB=90°-α, ∴∠AFB=90°-α.
∠AFB=90°+α的证明如下: ∵AB=AC,EC=ED,∠BAC=∠CED, ∴△ABC∽△EDC, ∴∠ACB=∠ECD,=, ∴∠BCD=∠ACE, ∴△BCD∽△ACE, ∴∠BDC=∠AEC, ∴∠AFB=∠BDC+∠CDE+∠DEF, =∠CDE+∠CED=180°-∠DCE, ∵AB=AC,EC=ED,∠BAC=∠DEC=α, ∴∠DCE=90°-α, ∴∠AFB=180°-(90°-α)=90°+α. |