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6. 已知在$□ ABCD$中,$AC=6$,$E$是$AD$上一点,$△ DCE$的周长是$□ ABCD$周长的一半,且$EC=4$,连接$EO$,则$EO$的长为
(
A. 3
B. 5
C. $2\sqrt{5}$
D. $\sqrt{7}$
(
D
)A. 3
B. 5
C. $2\sqrt{5}$
D. $\sqrt{7}$
答案
6. D
7. 如图,两条宽度分别为1和2的长方形纸条交叉放置,重叠部分为四边形$ABCD$,若$AB+BC=6$,则四边形$ABCD$的面积是
(
A. 4
B. 2
C. 8
D. 5
(
A
)A. 4
B. 2
C. 8
D. 5
答案
7. A
8. 如图,在$□ ABCD$中,$AE⊥ BC$于点$E$,$AF⊥ CD$于点$F$,若$∠ EAF=60°$,$BE=4$,$DF=6$,则$□ ABCD$的周长为
40
,面积为48√3
.答案
8. $40,48\sqrt{3}$
9. 如图,四边形$ABCD$是平行四边形,$AE⊥ BC$且交$CB$的延长线于点$E$,$DF⊥ BC$于点$F$.证明:$BE=CF$.
答案
9. 证明:$\because$四边形$ABCD$是平行四边形,
$\therefore AB// DC,AB=DC$,
$\therefore ∠ ABE=∠ C$,
$\because AE⊥ BC$且交$CB$的延长线于点$E$,$DF⊥ BC$于点$F$,
$\therefore ∠ E=∠ DFC=90°$,
在$△ ABE$和$△ DCF$中,
$\begin{cases}∠ E=∠ DFC, \\∠ ABE=∠ C, \\AB=DC,\end{cases}$
$\therefore △ ABE≌△ DCF(\mathrm{AAS})$,
$\therefore BE=CF$.
$\therefore AB// DC,AB=DC$,
$\therefore ∠ ABE=∠ C$,
$\because AE⊥ BC$且交$CB$的延长线于点$E$,$DF⊥ BC$于点$F$,
$\therefore ∠ E=∠ DFC=90°$,
在$△ ABE$和$△ DCF$中,
$\begin{cases}∠ E=∠ DFC, \\∠ ABE=∠ C, \\AB=DC,\end{cases}$
$\therefore △ ABE≌△ DCF(\mathrm{AAS})$,
$\therefore BE=CF$.
10. 如图,在$□ ABCD$中,$∠ ABC$为锐角,点$E$在边$AD$上,连接$BE$,$CE$,且$S_{△ ABE}=S_{△ CDE}$,若$F$是边$BC$的中点,连接$EF$,对角线$AC$与$EF$相交于点$H$.求证:点$H$是$AC$的中点.
答案
10. 证明:$\because$四边形$ABCD$为平行四边形,
$\therefore AD// BC,AD=BC$,
$\therefore AD$与$BC$之间是等距的,且$∠ EAH=∠ FCH$.
$\because S_{△ ABE}=S_{△ CDE}$,
$\therefore AE=DE=\dfrac{1}{2}AD$.
$\because F$为$BC$的中点,
$\therefore CF=BF=\dfrac{1}{2}BC$,
$\therefore CF=AE$.
在$△ AEH$和$△ CFH$中,$\begin{cases}∠ EHA=∠ FHC, \\∠ EAH=∠ FCH, \\AE=CF,\end{cases}$
$\therefore △ AEH≌△ CFH(\mathrm{AAS})$,
$\therefore AH=CH$,
$\therefore H$为$AC$的中点.
$\therefore AD// BC,AD=BC$,
$\therefore AD$与$BC$之间是等距的,且$∠ EAH=∠ FCH$.
$\because S_{△ ABE}=S_{△ CDE}$,
$\therefore AE=DE=\dfrac{1}{2}AD$.
$\because F$为$BC$的中点,
$\therefore CF=BF=\dfrac{1}{2}BC$,
$\therefore CF=AE$.
在$△ AEH$和$△ CFH$中,$\begin{cases}∠ EHA=∠ FHC, \\∠ EAH=∠ FCH, \\AE=CF,\end{cases}$
$\therefore △ AEH≌△ CFH(\mathrm{AAS})$,
$\therefore AH=CH$,
$\therefore H$为$AC$的中点.
