第146页

信息发布者:
①②

解:
(1)证明如下:
$\because DF// AE,$
$\therefore ∠ A=∠ DFB.$
$\because ∠ FDE=∠ A,$
$\therefore ∠ FDE=∠ DFB,$
$\therefore DE// BA.$
(2)$\because ∠ FDE=∠ A,$$∠ A=∠ BDF=2∠ EDC,$
$∠ FDE+∠ BDF+∠ EDC=180°,$
$\therefore ∠ A+∠ A+\frac{1}{2}∠ A=180°,$
解得$∠ A=72°.$
$\because DF// AE,$
$\therefore ∠ AFD=180°-∠ A=108°.$
解:
(1)$\because ∠ A=50°,$$∠ C=30°,$
$\therefore ∠ BDO=∠ A+∠ C=80°.$
$\because ∠ BOD=70°,$
$\therefore ∠ B=180°-∠ BDO-∠ BOD=180°-80°-70°=30°.$
(2)$∠ BOC=∠ A+∠ B+∠ C.$
证明:$\because ∠ BEC=∠ A+∠ B,$
$\therefore ∠ BOC=∠ BEC+∠ C=∠ A+∠ B+∠ C.$