登录
1. 如图13-3-9，将长方形纸片ABCD沿对角线BD折叠，点C的对应点为点E，BE交AD于点O。若∠CBD=31°，则∠BOD的度数为(
A.118°
B.111°
C.101°
D.62°
图13-3-9 图13-3-10 图13-3-11
A
)A.118°
B.111°
C.101°
D.62°
图13-3-9 图13-3-10 图13-3-11
答案
1.A
2.若△ABC三个角的大小满足∠A:∠B:∠C=2:3:4,则∠A的度数为(
A.40°
B.60°
C.80°
D.100°
A
)A.40°
B.60°
C.80°
D.100°
答案
2.A
3. 如图13-3-10，∠A+∠B+∠C+∠D+∠E+∠F一定等于（
A.$180°$
B.$360°$
C.$540°$
D.$720°$
B
）A.$180°$
B.$360°$
C.$540°$
D.$720°$
答案
3.B 点拨:$∠ A+∠ E+∠ C=180°,∠ D+∠ B+∠ F=180°$.
4.【教材P12例2变式】如图13-3-11,一轮船在海上往东行驶,在A处测得灯塔C位于北偏东$60°$方向,在B处测得灯塔C位于北偏东$25°$方向,则$∠ ACB$的度数为(
A.$45°$
B.$35°$
C.$25°$
D.$30°$
B
)A.$45°$
B.$35°$
C.$25°$
D.$30°$
答案
4.B
5. 如图13-3-12所示，在△ABC中，AD交BC于点D，∠ADB=100°，BE是△ABC的角平分线，BE，AD相交于点F，已知∠BAD=40°，求∠BFD的度数. 
答案
5. 解: $\because$ 在 $△ ABD$ 中, $∠ ADB = 100°$,$∠ BAD=40°$,
$\therefore ∠ ABD=180°-100°-40°=40°$.
$\because BE$ 是 $△ ABC$ 的角平分线,
$\therefore ∠ FBD=\frac{1}{2}∠ ABD=20°$,
$\therefore ∠ BFD=180°-100°-20°=60°$.
$\therefore ∠ ABD=180°-100°-40°=40°$.
$\because BE$ 是 $△ ABC$ 的角平分线,
$\therefore ∠ FBD=\frac{1}{2}∠ ABD=20°$,
$\therefore ∠ BFD=180°-100°-20°=60°$.
6. 如图13-3-13,在$△ ABC$中,$D$为$BC$上一点,将$△ ABD$沿$AD$翻折得到$△ AED$,$AE$与$BC$相交于点$F$,若$AE$平分$∠ CAD$,$∠ B=40°$,$∠ C=35°$.
(1)求证$∠ CAF=∠ C$;
(2)求$∠ 1$的度数.
(1)求证$∠ CAF=∠ C$;
(2)求$∠ 1$的度数.
答案
6. (1)证明: $\because ∠ B=40°,∠ C=35°$,
$\therefore ∠ BAC=180°-40°-35°=105°$.
$\because AE$ 平分 $∠ CAD$,
$\therefore ∠ CAF=∠ DAF$.
由翻折,可知$∠ BAD=∠ DAF$,
$\therefore ∠ CAF=∠ DAF=∠ BAD$,
$\therefore ∠ CAF=\frac{1}{3}∠ BAC=35°$,
$\therefore ∠ CAF=∠ C$.
(2)解: $\because ∠ CAF=∠ C=35°$,
$\therefore ∠ AFC=180°-35°-35°=110°$.
$\because ∠ AFC=∠ DFE,∠ E=∠ B=40°$,
$\therefore ∠ 1=180°-110°-40°=30°$.
$\therefore ∠ BAC=180°-40°-35°=105°$.
$\because AE$ 平分 $∠ CAD$,
$\therefore ∠ CAF=∠ DAF$.
由翻折,可知$∠ BAD=∠ DAF$,
$\therefore ∠ CAF=∠ DAF=∠ BAD$,
$\therefore ∠ CAF=\frac{1}{3}∠ BAC=35°$,
$\therefore ∠ CAF=∠ C$.
(2)解: $\because ∠ CAF=∠ C=35°$,
$\therefore ∠ AFC=180°-35°-35°=110°$.
$\because ∠ AFC=∠ DFE,∠ E=∠ B=40°$,
$\therefore ∠ 1=180°-110°-40°=30°$.
