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8.如图，$△ ABC$中，点$D$，$E$分别在$AB$，$AC$上，且$\frac{AD}{DB}=\frac{AE}{EC}=\frac{1}{2}$，下列结论正确的是 （
A.$DE:BC=1:2$
B.$△ ADE$与$△ ABC$的面积比为$1:3$
C.$△ ADE$与$△ ABC$的周长比为$1:2$
D.$DE// BC$
D
）A.$DE:BC=1:2$
B.$△ ADE$与$△ ABC$的面积比为$1:3$
C.$△ ADE$与$△ ABC$的周长比为$1:2$
D.$DE// BC$
答案
8.D
9.如图,在平行四边形ABCD中,E为CD上一点,DE:CE=2:3,连接AE,BD交于点F,则$S_{△ DEF}:S_{△ ADF}:S_{△ ABF}$等于（
A.$2:3:5$
B.$4:9:25$
C.$4:10:25$
D.$2:5:25$
C
）A.$2:3:5$
B.$4:9:25$
C.$4:10:25$
D.$2:5:25$
答案
9.C
10.某社区拟筹资金2000元，计划在一块上、下底分别是10 m，20 m的梯形空地上种植花木(如图)，他们想在△AMD和△BMC地带种植单价为10元/m²的太阳花，当△AMD地带种满花后，已经花了500元，请你预算一下，若继续在△BMC地带种植同样的太阳花，资金是否够用，并说明理由.
答案
10.解:资金不够用.理由:$S_{△ ADM} = 500 ÷ 10 = 50(\mathrm{m}^2).\because AD // BC$，
$\therefore △ ADM ∽ △ CBM, \therefore \frac{S_{△ ADM}}{S_{△ CBM}} = ( \frac{10}{20} )^2 = \frac{1}{4}, \therefore S_{△ CMB} = 4S_{△ ADM} = 4 × 50 = 200(\mathrm{m}^2), \therefore$ 还需资金 $10 × 200 = 2\ 000$(元),$\therefore$共需资金 $500+2\ 000=2\ 500$(元)$＞2\ 000$ 元,$\therefore$资金不够用.
$\therefore △ ADM ∽ △ CBM, \therefore \frac{S_{△ ADM}}{S_{△ CBM}} = ( \frac{10}{20} )^2 = \frac{1}{4}, \therefore S_{△ CMB} = 4S_{△ ADM} = 4 × 50 = 200(\mathrm{m}^2), \therefore$ 还需资金 $10 × 200 = 2\ 000$(元),$\therefore$共需资金 $500+2\ 000=2\ 500$(元)$＞2\ 000$ 元,$\therefore$资金不够用.
11.如图，在$△ ABC$中，点$D$，$E$，$F$分别在$AB$，$BC$，$AC$边上，$DE// AC$，$EF// AB$。
(1)求证：$△ BDE∽△ EFC$；
(2)设$\frac{AF}{FC}=\frac{1}{2}$。
①若$BC＝12$，求线段$BE$的长；
②若$△ EFC$的面积是$20$，求$△ ABC$的面积。
(1)求证：$△ BDE∽△ EFC$；
(2)设$\frac{AF}{FC}=\frac{1}{2}$。
①若$BC＝12$，求线段$BE$的长；
②若$△ EFC$的面积是$20$，求$△ ABC$的面积。
答案
11.(1) 证明: $\because DE // AC, \therefore ∠ DEB = ∠ FCE. \because EF // AB$,
$\therefore ∠ DBE = ∠ FEC, \therefore △ BDE ∽ △ EFC$.
(2)解: ①$\because EF // AB, \therefore \frac{BE}{EC} = \frac{AF}{FC} = \frac{1}{2}. \because EC = BC - BE = 12 - BE, \therefore \frac{BE}{12 - BE} = \frac{1}{2}, \therefore BE = 4$.
②$\because \frac{AF}{FC} = \frac{1}{2}, \therefore \frac{FC}{AC} = \frac{2}{3}. \because EF // AB, \therefore △ EFC ∽ △ BAC$,
$\therefore \frac{S_{△ EFC}}{S_{△ ABC}} = ( \frac{FC}{AC} )^2 = ( \frac{2}{3} )^2 = \frac{4}{9}, \therefore S_{△ ABC} = \frac{9}{4} S_{△ EFC} = \frac{9}{4} × 20 = 45$.
$\therefore ∠ DBE = ∠ FEC, \therefore △ BDE ∽ △ EFC$.
(2)解: ①$\because EF // AB, \therefore \frac{BE}{EC} = \frac{AF}{FC} = \frac{1}{2}. \because EC = BC - BE = 12 - BE, \therefore \frac{BE}{12 - BE} = \frac{1}{2}, \therefore BE = 4$.
②$\because \frac{AF}{FC} = \frac{1}{2}, \therefore \frac{FC}{AC} = \frac{2}{3}. \because EF // AB, \therefore △ EFC ∽ △ BAC$,
$\therefore \frac{S_{△ EFC}}{S_{△ ABC}} = ( \frac{FC}{AC} )^2 = ( \frac{2}{3} )^2 = \frac{4}{9}, \therefore S_{△ ABC} = \frac{9}{4} S_{△ EFC} = \frac{9}{4} × 20 = 45$.
12.如图,在$△ ABC$中,$∠ C=90°$,$AD$与$BD$分别是$△ ABC$的内角$∠ BAC$,$∠ ABC$的平分线,过点$A$作$AE⊥ AD$交$BD$的延长线于点$E$,$△ ABC∽△ EDA$.
(1)求$∠ ABC$的度数;
(2)求$\frac{S_{△ ABC}}{S_{△ EDA}}$的值.
(1)求$∠ ABC$的度数;
(2)求$\frac{S_{△ ABC}}{S_{△ EDA}}$的值.
答案
12.解:(1)$\because AD$ 与 $BD$ 分别是$△ ABC$ 的内角$∠ BAC,∠ ABC$ 的平分线, $\therefore ∠ 1 = \frac{1}{2} ∠ ABC, ∠ 2 = \frac{1}{2} ∠ BAC. \because ∠ C = 90°$,
$\therefore ∠ 1 + ∠ 2 = \frac{1}{2} (∠ ABC + ∠ BAC) = \frac{1}{2} × 90° = 45°, \therefore ∠ 3 = ∠ 1 + ∠ 2 = 45°. \because △ ABC ∽ △ EDA, \therefore ∠ ABC = ∠ 3 = 45°$.
(2)过 $A$ 作 $AF ⊥ DE$ 于点 $F$.
$\because ∠ 3 = 45°, AE ⊥ AD, \therefore △ ADE$ 是等腰直角三角形.设 $AF = a$, 则 $DE = 2a, DF = a$, Rt$△ ADF$ 中, $AD = \sqrt{2}\ a. \because 2∠ 1 = 2∠ 2 = 45°$,
$\therefore ∠ 1 = ∠ 2, \therefore AD = BD = \sqrt{2}\ a, \therefore BF = \sqrt{2}\ a + a$. 在 Rt$△ ABF$ 中, $AB^2 = AF^2 + BF^2 = a^2 + (\sqrt{2}\ a + a)^2 = (4 + 2\sqrt{2}\ )a^2$.
$\because △ ABC ∽ △ EDA, \therefore \frac{S_{△ ABC}}{S_{△ EDA}} = \frac{AB^2}{ED^2} = \frac{(4 + 2\sqrt{2}\ )a^2}{(2a)^2} = \frac{2 + \sqrt{2}}{2}$.
