∴△ABC∽△EDC,
∴∠CBD=∠CAE,
∴∠AFB=180°-∠CAE-∠BAC-∠ABD
=180°-∠BAC-∠ABC
=∠ACB,
∴∠AFB=60°,
同理可得:∠AFB=45°,
故答案為:60°,45°;
(2)∠AFB=90°?
1 |
2 |
證明:∵AB=AC,EC=ED,∠BAC=∠CED
∴△ABC∽△EDC,
∴
BC |
DC |
AC |
EC |
∵∠BCD=∠ACE,
∴△BCD∽△ACE,
∴∠CBD=∠CAE.
∠AFB=∠CBD+∠AEC=∠CAE+∠AEC=∠ACB.
∵AB=AC,∠BAC=a,
∴∠ACB=90°?
1 |
2 |
∴∠AFB=90°?
1 |
2 |