10.如图,在$△ ABC$中,$AB=AC$,$AD⊥ BC$,$CE⊥ AB$,$EF=BE$.
(1)$△ AEF$与$△ CEB$全等吗?说明理由;
(2)说明$AF=2BD$的理由.

(1)$△ AEF$与$△ CEB$全等吗?说明理由;
(2)说明$AF=2BD$的理由.
答案
10. (1)解:全等. 理由: $\because AD ⊥ BC, \therefore ∠ B+∠ EAF=90°. \because CE ⊥ AB, \therefore ∠ B + ∠ ECB = 90°, ∠ AEF = ∠ CEB = 90°. \therefore ∠ EAF = ∠ ECB.$
又$\because EF = BE, \therefore △ AEF ≌ △ CEB$(AAS).
(2) $\because △ AEF ≌ △ CEB, \therefore AF=CB. \because AB=AC, AD ⊥ BC, \therefore BC=2BD. \therefore AF=2BD.$
又$\because EF = BE, \therefore △ AEF ≌ △ CEB$(AAS).
(2) $\because △ AEF ≌ △ CEB, \therefore AF=CB. \because AB=AC, AD ⊥ BC, \therefore BC=2BD. \therefore AF=2BD.$
11. 如图所示的平面直角坐标系中,点A的坐标为(4,2),点B的坐标为(1,-3),在y轴上取一点P,使PA+PB的值最小,则点P的坐标为(

A.(2,0)
B.(-2,0)
C.(0,2)
D.(0,-2)
D
)A.(2,0)
B.(-2,0)
C.(0,2)
D.(0,-2)
答案
11.D
12. 在$△ ABC$中,$AB=AC$,$∠ BAC=100°$,点$D$在$BC$边上,连接$AD$。若$△ ABD$为直角三角形,则$∠ ADC$的度数为$\underline{\hspace{5cm}}$。
答案
12. $90° 或 130°$
13. 如图,在△ABC中,AB=AC,∠B=70°,以点C为圆心,CA长为半径作弧,交直线BC于点P,连接AP,则∠BAP的度数是

15°
或75°
.答案
13. $15° 或 75°$
14.如图,等边三角形ABC和等边三角形A'B'C的边长都是4,点B,C,B'在同一条直线上,点P在线段A'C上,则$AP+BP$的最小值为

8
。答案
14.8
15.在$△ ABC$中,$BC=10\mathrm{cm}$,$AB$的垂直平分线与$AC$的垂直平分线分别交线段$BC$于点$D$,$E$,$DE=2\mathrm{cm}$,则$AD+AE=\_\_\_\_\_\_\mathrm{cm}.$
答案
15. 8 或 12
16.(仙桃市期末)如图所示,在$△ ABC$中,$AB=AC=2$,$∠ B=40°$,点$D$在线段$BC$上运动(点$D$不与点$B$,$C$重合),连接$AD$,作$∠ ADE=40°$,$DE$交线段$AC$于点$E$.
(1)当$∠ BDA=115°$时,$∠ BAD=\_\_\_\_\_\_$;点$D$沿$BC$方向运动时,$∠ BDA$逐渐变________(填“大”或“小”).
(2)当$DC$的长为多少时,$△ ABD$与$△ DCE$全等?请说明理由.
(3)在点$D$的运动过程中,$△ ADE$的形状也在改变,请判断当$∠ BDA$等于多少度时,$△ ADE$是等腰三角形.(直接写出结论,不用说明理由)

(1)当$∠ BDA=115°$时,$∠ BAD=\_\_\_\_\_\_$;点$D$沿$BC$方向运动时,$∠ BDA$逐渐变________(填“大”或“小”).
(2)当$DC$的长为多少时,$△ ABD$与$△ DCE$全等?请说明理由.
(3)在点$D$的运动过程中,$△ ADE$的形状也在改变,请判断当$∠ BDA$等于多少度时,$△ ADE$是等腰三角形.(直接写出结论,不用说明理由)
答案
16. (1)$25°$ 小
(2)解: 当$DC=2$时,$△ ABD ≌ △ DCE$. 理由: $\because AB=2$,$DC=2, \therefore AB=DC, \because AB=AC, \therefore ∠ C = ∠ B = 40°, \therefore ∠ DEC + ∠ EDC = 140°. \because ∠ ADE=40°, \therefore ∠ ADB + ∠ EDC = 140°, \therefore ∠ ADB = ∠ DEC.$ 在$△ ABD$和$△ DCE$中,
$\begin{cases}∠ ADB = ∠ DEC,\\∠ B = ∠ C,\\AB = DC,\end{cases}$
$\therefore △ ABD ≌ △ DCE(\mathrm{AAS}).$
(3)当$∠ BDA$的度数为$110°$或$80°$时,$△ ADE$是等腰三角形.
(2)解: 当$DC=2$时,$△ ABD ≌ △ DCE$. 理由: $\because AB=2$,$DC=2, \therefore AB=DC, \because AB=AC, \therefore ∠ C = ∠ B = 40°, \therefore ∠ DEC + ∠ EDC = 140°. \because ∠ ADE=40°, \therefore ∠ ADB + ∠ EDC = 140°, \therefore ∠ ADB = ∠ DEC.$ 在$△ ABD$和$△ DCE$中,
$\begin{cases}∠ ADB = ∠ DEC,\\∠ B = ∠ C,\\AB = DC,\end{cases}$
$\therefore △ ABD ≌ △ DCE(\mathrm{AAS}).$
(3)当$∠ BDA$的度数为$110°$或$80°$时,$△ ADE$是等腰三角形.
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