7.若两个相似三角形对应边的比为$1:8$,则这两个相似三角形周长的比为()

A.$1:2$
B.$1:4$
C.$1:8$
D.$1:16$

7. C

8. 如图,若$△ ABC$与$△ DEF$相似,则它们的相似比是()

A.$\sqrt{2}:1$
B.$2:1$
C.$2:\sqrt{3}$
D.$\sqrt{5}:2$

8. A 解析 $\because △ ABC ∽ △ DEF,\therefore$两个三角形最短边的比等于这两个三角形的相似比,即$\frac{2}{\sqrt{2}}=\sqrt{2}:1.$

9.如图,在$△ ABC$中,延长$AC$至点$D$,使$CD=CA$,过点$D$作$DE// CB$,且$DE=DC$,连接$AE$交$BC$于点$F$.若$∠ CAB=∠ CFA$,$CF=1$,则$BF=$.

9. 3 解析 $\because CD=CA,DE// CB,\therefore AF=EF,\therefore CF$是$△ ADE$的中位线,$\therefore DE=2CF=2. \because DE=DC=AC,\therefore AC=2. \because ∠ CAB=∠ CFA,∠ ACF=∠ BCA,\therefore △ CAF ∽ △ CBA,\therefore AC:BC=CF:CA,\therefore 2:BC=1:2,\therefore BC=4,\therefore BF=BC-CF=3.$

10. 如图,在$△ ABC$和$△ DEC$中,$∠ A=∠ D$,$∠ BCE=∠ ACD$.
(1)求证:$△ ABC ∽ △ DEC$.
(2)若$S_{△ ABC}:S_{△ DEC}=4:9$,$BC=6$,求$CE$的长.

10. (1)证明:$\because ∠ BCE=∠ ACD,\therefore ∠ BCE+∠ ACE=∠ ACD+∠ ACE, \therefore ∠ ACB = ∠ DCE.$又$\because ∠ A=∠ D,\therefore △ ABC ∽ △ DEC.$
(2)解:由(1)知$△ ABC ∽ △ DEC,\therefore \frac{S_{△ ABC}}{S_{△ DEC}}=(\frac{CB}{CE})^2=\frac{4}{9},\therefore \frac{CB}{CE}=\frac{2}{3}.$又$\because BC=6,\therefore CE=9.$

11. 如图,在梯形 $ABCD$ 中,$AD // BC$,点 $F,E$ 分别在线段 $BC,AC$ 上,且 $∠ FAC = ∠ ADE$, $AC=AD$.
(1) 求证: $AF=DE$.
(2) 若 $∠ ABC = ∠ CDE$, 求证: $AF^2 = BF · CE$.

证明:(1) $\because AD // BC, \therefore ∠ ACF = ∠ DAC.$
$\because ∠ FAC = ∠ ADE, AC = AD, \therefore △ ACF ≌ △ DAE(\mathrm{ASA}), \therefore AF = DE.$
(2) 由(1)知$△ ACF ≌ △ DAE, \therefore ∠ AFC = ∠ DEA, \therefore ∠ AFB = ∠ DEC. \because ∠ ABC = ∠ CDE, \therefore △ ABF ∽ △ CDE, \therefore \frac{AF}{CE}=\frac{BF}{DE}, \therefore AF · DE = BF · CE.$
$\because AF=DE,\therefore AF^2=BF · CE.$