答案：7. A

答案：8. $89°$

答案：
9.解:(1)$\because AC// DE$,$\therefore ∠ D+∠ ACD=180°$.
又$\because ∠ D+∠ BAC=180°$,
$\therefore ∠ ACD=∠ BAC$,$\therefore AB// CD$.
(2)如答图.
$\because AC// DE$,$∠ CED=35°$,
$\therefore ∠ ACE=∠ CED=35°$.

$\because CE$平分$∠ ACD$,$\therefore ∠ ACD=2∠ ACE=70°$.
由(1)知$AB// CD$,$\therefore ∠ BAC=∠ ACD=70°$.
又$\because AB⊥ BC$,$\therefore ∠ B=90°$,
$\therefore ∠ ACB=180°-∠ B-∠ BAC=180°-90°-70°=20°$.

答案：
10.解:(1)$\because AE// BD$,$\therefore ∠ A+∠ ABD=180°$.
$\because ∠ A=∠ BDC$,$\therefore ∠ BDC+∠ ABD=180°$,
$\therefore AB// CD$.
(2)$∠ A+∠ AEC+∠ C=360°$.理由如下:
如答图,过点$E$作$EH// AB$.
由(1)知$AB// CD$,$\therefore AB// EH// CD$,
$\therefore ∠ A+∠ AEH=180°$,$∠ C+∠ CEH=180°$,
$\therefore ∠ A+∠ AEH+∠ C+∠ CEH=360°$,
即$∠ A+∠ AEC+∠ C=360°$.

(3)$\because ∠ AEC$的平分线交$CD$的延长线于点$F$,
$\therefore ∠ CEF=\frac{1}{2}∠ AEC$.
$\because$在三角形$CEF$中,$∠ F+∠ CEF+∠ C=180°$,
$∠ F=20°$,
$\therefore \frac{1}{2}∠ AEC+∠ C=160°$①.
$\because ∠ A=∠ BDC$,$∠ BDC=140°$,$\therefore ∠ A=140°$.
$\because ∠ A+∠ AEC+∠ C=360°$,
$\therefore ∠ AEC+∠ C=220°$②.
②$-$①,得$\frac{1}{2}∠ AEC=60°$,$\therefore ∠ AEC=120°$,
$\therefore ∠ C=100°$.