$解：(1)∵BF//AD,∴∠CED=∠CFB,∠CBF=∠D.$
$∵∠ADB= \frac{1}{2} ∠ABC,∠ABC=∠ABF+∠CBF. $
$∴∠ABF=∠CBF= \frac{1}{2} ∠ABC. $
$∵CE是∠ACB的平分线,∴∠FCB= \frac{1}{2} ∠ACB,$
$ \begin{aligned}∴∠CED&=∠CFB \\ &=180°-(∠FCB+∠FBC) \\ &=180°- \frac{1}{2} (∠ACB+∠ABC) \\ &=180°- \frac{1}{2} (180°-∠BAC) \\ &=130°. \\ \end{aligned}$
$（更多请点击查看作业精灵详解）$
