9.(凉山州中考)如图，△ABC中，∠BCD=30°，∠ACB=80°，CD是边AB上的高，AE是∠CAB的平分线，则∠AEB的度数是。

9.$100°$

10.【方程思想】如图，AD平分∠BAC，∠EAD=∠EDA.
(1)求证:∠EAC=∠B;
(2)若∠B=50°,∠CAD:∠E=1:3,求∠E的度数.

10. (1)证明:$\because AD$平分$∠ BAC, \therefore ∠ BAD = ∠ CAD$. 又$\because ∠ EAD = ∠ EDA, \therefore ∠ EAC = ∠ EAD - ∠ CAD = ∠ EDA - ∠ BAD = ∠ B$,
$\therefore ∠ EAC = ∠ B$.
(2)解: 设$∠ CAD = x°$, 则$∠ E = 3x°$. 由(1)知$∠ EAC = ∠ B = 50°, \therefore ∠ EAD = ∠ EDA = (x+50)°$. 在$△ EAD$中,
$∠ E + ∠ EAD + ∠ EDA = 180°, \therefore 3x° + 2(x+50)° = 180°$, 解得$x=16. \therefore ∠ E = 48°$.

11.等腰三角形的一个外角为110°,则它的底角为（）

A.55°
B.70°
C.55°或70°
D.以上都不对

11.C

12. 如图是四条互相不平行的直线$l_1,l_2,l_3,l_4$所截出的七个角，关于这七个角的度数关系，下列结论中正确的是（）

A.$∠2=∠4+∠7$
B.$∠3=∠1+∠7$
C.$∠1+∠4+∠6=180°$
D.$∠2+∠3+∠5=360°$

12.B

13.在$△ ABC$中，$∠ A=50°$，$∠ B=30°$，点D在AB边上，连接CD.若$△ ACD$为直角三角形，则$∠ BCD$的度数为.

13.$60°$或$10°$

14.【一题多设问】如图,在$△ ABC$中,$∠ ABC$与$∠ ACB$的平分线交于点$D$,$DE⊥ BC$于点$E$,$∠ BDE - ∠ DCE = n$.
(1)若$∠ ABC=60°,∠ ACB=40°$,求$n$的大小；
(2)若$∠ A=60°$,求$n$的大小；
(3)若$∠ A=α$,试用含$α$的式子表示$n$.

14. (1)解:$\because BD, CD$分别平分$∠ ABC, ∠ ACB, \therefore ∠ DCE = \frac{1}{2}∠ ACB = \frac{1}{2} × 40° = 20°, ∠ DBE = \frac{1}{2}∠ ABC = \frac{1}{2} × 60° = 30°. \because DE ⊥ BC, \therefore ∠ BDE = 90° - 30° = 60°. \therefore ∠ BDE - ∠ DCE = 60° - 20° = 40°$, 即$n=40°$.
(2)$\because BD, CD$分别平分$∠ ABC, ∠ ACB, \therefore$设$∠ DCE = ∠ ACD = x, ∠ DBE = ∠ ABD = y$, 则$∠ ACB = 2x, ∠ ABC = 2y. \because ∠ A = 60°, \therefore 2x + 2y = 120°, \therefore x + y = 60°. \because DE ⊥ BC, \therefore ∠ BDE = 90° - y. \therefore ∠ BDE - ∠ DCE = 90° - y - x = 90° - 60° = 30°$, 即$n=30°$.
(3)同(2)设$∠ DCE = ∠ ACD = p, ∠ DBE = ∠ ABD = q$, 则$∠ ACB = 2p, ∠ ABC = 2q. \because ∠ A = α, \therefore 2p + 2q = 180° - α. \therefore p + q = 90° - \frac{1}{2}α. \because DE ⊥ BC, \therefore ∠ BDE = 90° - q. \therefore ∠ BDE - ∠ DCE = 90° - q - p = 90° - (90° - \frac{1}{2}α) = \frac{1}{2}α$, 即$n = \frac{1}{2}α$.