7. (2025·达州模拟)如图，将一张含有$30°$角的三角形纸片的两个顶点叠放在长方形的两条对边上.若$∠ 2=44°$，则$∠ 1$的度数为 (
A.$14°$
B.$44°$
C.$46°$
D.$74°$
A
)A.$14°$
B.$44°$
C.$46°$
D.$74°$
答案：7. A
8. (2024·房山区期末)如图，$AB// CD$，$∠ BAE=123°$，$∠ DCE=32°$，则$∠ AEC$的度数为
$89°$
.答案：8. $89°$
9. 如图，已知$AC// DE$，$∠ D+∠ BAC=180°$.
(1)试说明$AB// CD$；
(2)连接$CE$，恰好满足$CE$平分$∠ ACD$，若$AB⊥ BC$，$∠ CED=35°$，求$∠ ACB$的度数.
(1)试说明$AB// CD$；
(2)连接$CE$，恰好满足$CE$平分$∠ ACD$，若$AB⊥ BC$，$∠ CED=35°$，求$∠ ACB$的度数.
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$\because AB⊥ BC$,$\therefore ∠ B=90°$,
$\therefore ∠ ACB=180°-∠ B-∠ BAC=180°-90°-70°=20°$.
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$\because AB⊥ BC$,$\therefore ∠ B=90°$,
$\therefore ∠ ACB=180°-∠ B-∠ BAC=180°-90°-70°=20°$.
10. (2024·海安期末)如图，$AE// BD$，$∠ A=∠ BDC$，$∠ AEC$的平分线交$CD$的延长线于点$F$.
(1)试说明$AB// CD$；
(2)探究$∠ A$，$∠ AEC$，$∠ C$之间的数量关系，并说明理由；
(3)若$∠ BDC=140°$，$∠ F=20°$，求$∠ C$的度数.
(1)试说明$AB// CD$；
(2)探究$∠ A$，$∠ AEC$，$∠ C$之间的数量关系，并说明理由；
(3)若$∠ BDC=140°$，$∠ F=20°$，求$∠ C$的度数.
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$\frac{1}{2}∠ AEC=60°$,$\therefore ∠ AEC=120°$,
$\therefore ∠ C=100°$.
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$\frac{1}{2}∠ AEC=60°$,$\therefore ∠ AEC=120°$,
$\therefore ∠ C=100°$.