(2)=1.
图1-3-1
(1)证法一:∵EF∥BC,
∴∠AFE=∠C.
又∵DF∥AB,
∴∠A=∠DFC.
∴△AEF∽△FDC.
证法二:∵EF∥BC,∴△AEF∽△ABC.
又∵DF∥AB,∴△ABC∽△FDC.
(2)思路分析
证明:∵DF∥AB,
∴=.
又∵EF∥BC,∴=.
∴=+
==1.