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