8. (2025·威海期末)如图,$AE,DE$分别是四边形$ABCD$的外角$∠ NAD$,$∠ MDA$的平分线.若$∠ B=90°,∠ E=60°$,则$∠ C$的度数为
$150°$
.答案：8. $150°$
9. 请用两种方法证明:四边形的外角和为$360°$.(要求:结合图形,写出已知、求证,并证明)
答案：
9.解:已知:如答图,四边形ABCD中,$∠ 1$,$∠ 2$,$∠ 3$,$∠ 4$是
它的四个外角.
求证:$∠ 1+∠ 2+∠ 3+∠ 4=360°$.
证法一:$\because ∠ 1=180°-∠ BAD$,$∠ 2=180°-∠ ADC$,
$∠ 3=180°-∠ DCB$,$∠ 4=180°-∠ ABC$,
$\therefore ∠ 1+∠ 2+∠ 3+∠ 4=180°× 4-(∠ ABC+∠ BCD+$
$∠ CDA+∠ DAB)$.
$\because$四边形的内角和为$360°$,
$\therefore ∠ ABC+∠ BCD+∠ CDA+∠ DAB=360°$.
$\therefore ∠ 1+∠ 2+∠ 3+∠ 4=180°× 4-360°=360°$.
证法二:如答图,连接AC,BD.
$\because ∠ 1=∠ ABD+∠ ADB$,$∠ 3=∠ CBD+∠ CDB$,
$\therefore ∠ 1+∠ 3=∠ ABC+∠ ADC$.
同理$∠ 2+∠ 4=∠ DAB+∠ DCB$.
$\therefore ∠ 1+∠ 2+∠ 3+∠ 4=∠ ABC+∠ ADC+∠ DAB+$
$∠ DCB$.
$\because$四边形的内角和为$360°$,
$\therefore ∠ ABC+∠ BCD+∠ CDA+∠ DAB=360°$.
$\therefore ∠ 1+∠ 2+∠ 3+∠ 4=360°$.
9.解:已知:如答图,四边形ABCD中,$∠ 1$,$∠ 2$,$∠ 3$,$∠ 4$是
它的四个外角.
求证:$∠ 1+∠ 2+∠ 3+∠ 4=360°$.
证法一:$\because ∠ 1=180°-∠ BAD$,$∠ 2=180°-∠ ADC$,
$∠ 3=180°-∠ DCB$,$∠ 4=180°-∠ ABC$,
$\therefore ∠ 1+∠ 2+∠ 3+∠ 4=180°× 4-(∠ ABC+∠ BCD+$
$∠ CDA+∠ DAB)$.
$\because$四边形的内角和为$360°$,
$\therefore ∠ ABC+∠ BCD+∠ CDA+∠ DAB=360°$.
$\therefore ∠ 1+∠ 2+∠ 3+∠ 4=180°× 4-360°=360°$.
证法二:如答图,连接AC,BD.
$\because ∠ 1=∠ ABD+∠ ADB$,$∠ 3=∠ CBD+∠ CDB$,
$\therefore ∠ 1+∠ 3=∠ ABC+∠ ADC$.
同理$∠ 2+∠ 4=∠ DAB+∠ DCB$.
$\therefore ∠ 1+∠ 2+∠ 3+∠ 4=∠ ABC+∠ ADC+∠ DAB+$
$∠ DCB$.
$\because$四边形的内角和为$360°$,
$\therefore ∠ ABC+∠ BCD+∠ CDA+∠ DAB=360°$.
$\therefore ∠ 1+∠ 2+∠ 3+∠ 4=360°$.
10. 如图,在四边形$ABCD$中,$AD⊥ DC,BC⊥ AB$,$AE$平分$∠ BAD$,$CF$平分$∠ DCB$,$AE$交$CD$于点$E$,$CF$交$AB$于点$F$.
(1)求$∠ BAE+∠ DCF$的度数;
(2)证明:$AE// CF$.
(1)求$∠ BAE+∠ DCF$的度数;
(2)证明:$AE// CF$.
答案：10.(1)解:$\because AD⊥ DC$,$BC⊥ AB$,
$\therefore ∠ B=∠ D=90°$.
$\because$四边形ABCD的内角和为$360°$,
$\therefore ∠ DAB+∠ DCB=180°$.
$\because$AE平分$∠ BAD$,CF平分$∠ DCB$,
$\therefore ∠ DAE=∠ BAE=\frac{1}{2}∠ DAB$,$∠ BCF=∠ DCF=$
$\frac{1}{2}∠ DCB$.
$\therefore ∠ BAE+∠ DCF=\frac{1}{2}(∠ DAB+∠ DCB)=90°$.
(2)证明:$\because ∠ DAE+∠ DEA=90°$,$∠ DAE=∠ BAE$,
$\therefore ∠ BAE+∠ DEA=90°$.
由(1)知,$∠ BAE+∠ DCF=90°$,
$\therefore ∠ DEA=∠ DCF$.$\therefore AE// CF$.
$\therefore ∠ B=∠ D=90°$.
$\because$四边形ABCD的内角和为$360°$,
$\therefore ∠ DAB+∠ DCB=180°$.
$\because$AE平分$∠ BAD$,CF平分$∠ DCB$,
$\therefore ∠ DAE=∠ BAE=\frac{1}{2}∠ DAB$,$∠ BCF=∠ DCF=$
$\frac{1}{2}∠ DCB$.
$\therefore ∠ BAE+∠ DCF=\frac{1}{2}(∠ DAB+∠ DCB)=90°$.
(2)证明:$\because ∠ DAE+∠ DEA=90°$,$∠ DAE=∠ BAE$,
$\therefore ∠ BAE+∠ DEA=90°$.
由(1)知,$∠ BAE+∠ DCF=90°$,
$\therefore ∠ DEA=∠ DCF$.$\therefore AE// CF$.
11. (2025·赣州期中)四边形$ABCD$中,$∠ A=140°,∠ D=80°$.
(1)如图①,若$∠ B=∠ C$,试求出$∠ C$的度数;
(2)如图②,若$∠ ABC$的平分线$BE$交$DC$于点$E$,且$BE// AD$,试求出$∠ C$的度数;
(3)如图③,若$∠ ABC$和$∠ BCD$的平分线交于点$E$,试求出$∠ BEC$的度数.
(1)如图①,若$∠ B=∠ C$,试求出$∠ C$的度数;
(2)如图②,若$∠ ABC$的平分线$BE$交$DC$于点$E$,且$BE// AD$,试求出$∠ C$的度数;
(3)如图③,若$∠ ABC$和$∠ BCD$的平分线交于点$E$,试求出$∠ BEC$的度数.
答案：11.(1)$\because ∠ A=140°$,$∠ D=80°$,$∠ B=∠ C$,
$\therefore 140°+80°+2∠ C=360°$,解得$∠ C=70°$.
(2)$\because ∠ A=140°$,$∠ D=80°$,$BE// AD$,
$\therefore ∠ ABE=180°-∠ A=180°-140°=40°$,
$∠ BED=180°-∠ D=180°-80°=100°$.
$\because$BE是$∠ ABC$的平分线,
$\therefore ∠ EBC=∠ ABE=40°$.
$\therefore ∠ C=∠ DEB-∠ EBC=100°-40°=60°$.
(3)$\because ∠ A=140°$,$∠ D=80°$,
$\therefore ∠ ABC+∠ BCD=360°-∠ A-∠ D=140°$.
$\because$BE,CE分别平分$∠ ABC$,$∠ BCD$,
$\therefore ∠ EBC+∠ ECB=\frac{1}{2}(∠ ABC+∠ BCD)=\frac{1}{2}×$
$140°=70°$.
$\therefore ∠ BEC=180°-(∠ EBC+∠ ECB)=180°-70°=110°$.
$\therefore 140°+80°+2∠ C=360°$,解得$∠ C=70°$.
(2)$\because ∠ A=140°$,$∠ D=80°$,$BE// AD$,
$\therefore ∠ ABE=180°-∠ A=180°-140°=40°$,
$∠ BED=180°-∠ D=180°-80°=100°$.
$\because$BE是$∠ ABC$的平分线,
$\therefore ∠ EBC=∠ ABE=40°$.
$\therefore ∠ C=∠ DEB-∠ EBC=100°-40°=60°$.
(3)$\because ∠ A=140°$,$∠ D=80°$,
$\therefore ∠ ABC+∠ BCD=360°-∠ A-∠ D=140°$.
$\because$BE,CE分别平分$∠ ABC$,$∠ BCD$,
$\therefore ∠ EBC+∠ ECB=\frac{1}{2}(∠ ABC+∠ BCD)=\frac{1}{2}×$
$140°=70°$.
$\therefore ∠ BEC=180°-(∠ EBC+∠ ECB)=180°-70°=110°$.