1. 给出下列各组线段,其中是成比例线段的是 ()

A.$a=2\ \mathrm{cm},b=4\ \mathrm{cm},c=6\ \mathrm{cm},d=8\ \mathrm{cm}$
B.$a=\frac{1}{2}\ \mathrm{m},b=\frac{1}{4}\ \mathrm{m},c=\frac{1}{6}\ \mathrm{m},d=\frac{1}{8}\ \mathrm{m}$
C.$a=\sqrt{2}\ \mathrm{cm},b=\sqrt{3}\ \mathrm{dm},c=\sqrt{10}\ \mathrm{cm},d=2\sqrt{5}\ \mathrm{dm}$
D.$a=2\ \mathrm{dm},b=\sqrt{5}\ \mathrm{dm},c=2\sqrt{3}\ \mathrm{dm},d=\sqrt{15}\ \mathrm{dm}$

1.D

2. 若$\frac{a}{b}=\frac{c}{d}$，且$m≠0$，则下列比例式中成立的是（）

A.$\frac{a}{b+m}=\frac{c}{d+m}$
B.$\frac{a - m}{b}=\frac{c - m}{d}$
C.$\frac{a + m}{b + m}=\frac{c + m}{d + m}$
D.$\frac{a + bm}{b}=\frac{c + dm}{d}$

2.D

3.若$\frac{x}{6}=\frac{y}{4}=\frac{z}{3}(x,y,z均不为0),则\frac{x+3y}{3y-2z}=$

3.3

4. 如图,在$△ ABC$中,D为AC的中点,点E在BC上,且$BE=3CE$,AE,BD交于点F,则$\frac{AF}{EF}$的值为.

4.$\frac{4}{3}$ 【点拨】过点E作$EG// BD$,交AC于点G. $\because EG// BD,\therefore \frac{CE}{BC}=\frac{CG}{CD}.\because BE=3CE,\therefore BC=4CE. \therefore \frac{CG}{CD}=\frac{CE}{BC}=\frac{1}{4}.\because D$为AC的中点,$\therefore \frac{AD}{DG}=\frac{4}{3}.\because EG// BD,\therefore \frac{AF}{EF}=\frac{AD}{DG}=\frac{4}{3}.$

5. 如图,在矩形$ABCD$中,$AB＜BC$,点$E,F$分别在$CD,AD$边上,且$△ BCE$与$△ BFE$关于直线$BE$对称.点$G$在$AB$边上,$GC$分别与$BF,BE$交于$P$,$Q$两点.若$\frac{AB}{BC}=\frac{4}{5}$,$CE=CQ$,则$\frac{GQ}{CQ}=$.

5.$\frac{3}{2}$ 【点拨】如图，连结FQ. $\because$ 四边形ABCD是矩形，$\therefore AB// CD,∠ BAF=90°,BC=AD.\because \frac{AB}{BC}=\frac{4}{5},\therefore$ 设$AB=4a,BC=5a.\because △ BCE$与$△ BFE$关于直线BE对称，$\therefore BF=BC=5a,CQ=FQ,CE=FE,\therefore AF=\sqrt{BF^2-AB^2}=\sqrt{(5a)^2-(4a)^2}=3a. \therefore DF=AD-AF=5a-3a=2a.\because CQ=CE,\therefore CQ=FQ=FE=CE.\therefore$ 四边形CQFE是菱形.$\therefore FQ// CE.\therefore AB// FQ// CE.\therefore \frac{GQ}{CQ}=\frac{AF}{DF}=\frac{3a}{2a}=\frac{3}{2}.$

6. ★★ [广州越秀区期中] 如图,在菱形ABCD中,∠ABC=60°,AB=a,点E,F是对角线BD上的点(点E,F不与B,D重合),分别连结AE,EC,AF,CF,若四边形AECF是菱形,且与菱形ABCD是相似图形,那么菱形AECF的边长是.(用含a的代数式表示)

6.$\frac{\sqrt{3}}{3}a$ 【点拨】如图，连结AC，交BD于点O. $\because$ 四边形ABCD是菱形，$\therefore AB=BC=a,AC⊥ BD,OC=OA.\because ∠ ABC=60°,\therefore △ ABC$为等边三角形，$\therefore AC=AB=a,\therefore OC=OA=\frac{1}{2}AC=\frac{1}{2}a.\because$ 四边形AECF是菱形，$\therefore CE=CF,OE=OF.\because$ 菱形ABCD与菱形ECFA相似，$\therefore ∠ ECF=∠ ABC=60°,\therefore △ CEF$为等边三角形，$\therefore EF=CF,\therefore OE=OF=\frac{1}{2}CE.\because CO^2+EO^2=CE^2,\therefore (\frac{1}{2}a)^2+(\frac{1}{2}CE)^2=CE^2,$ 解得$CE=\frac{\sqrt{3}}{3}a$(负值已舍去).

7. 如图,已知$△ ABC$,$△ DCE$,$△ FEG$是三个全等的等腰三角形,底边$BC$,$CE$,$EG$在同一直线上,且$AB=3$,$BC=\sqrt{3}$,连结$BF$,分别交$AC$,$DC$,$DE$于点$P$,$Q$,$R$.有下列结论:
①$△ BFG ∽ △ ABC$;②$BQ=FQ$;
③$AP=2PC$;④$FE$平分$∠ BFG$.
其中正确的有 ()

A.1个
B.2个
C.3个
D.4个

7.C 【点拨】$\because △ ABC,△ DCE,△ FEG$是三个全等的等腰三角形，$\therefore BC=CE=EG=\sqrt{3},FG=AC=AB=3.\therefore BG=3\sqrt{3}.\therefore \frac{BG}{FG}=\sqrt{3}=\frac{FG}{EG}.$ 又$\because ∠ G=∠ G,\therefore △ BFG∽ △ FEG.\therefore △ BFG∽ △ ABC.$ 故①正确；由题意得$∠ FEG=∠ DCE,\therefore CD// EF.\therefore \frac{BC}{CE}=\frac{BQ}{QF}.\because BC=CE,\therefore BQ=FQ,$ 故②正确；易知$AC// FG,\therefore △ BPC∽ △ BFG.\therefore \frac{PC}{FG}=\frac{BC}{BG}=\frac{1}{3}.\because AC=FG,\therefore \frac{PC}{AC}=\frac{1}{3}.\therefore AP=2PC,$ 故③正确；$\because △ BFG∽ △ FEG,\therefore ∠ EBF=∠ EFG.\because EF=AB=3,BE=BC+CE=2\sqrt{3},\therefore EF≠ BE.\therefore ∠ BFE≠ ∠ EBF.\therefore ∠ EFG≠ ∠ BFE,$ 故④错误. 故选C.

8. 新视角最值探究题 如图，四边形ABCD是边长为5的正方形，点P是BD上一动点，以AP为斜边在AP边的右侧作等腰直角三角形APQ，∠AQP=90°，连结DQ，CQ，则DQ的最小值为。

8.$\frac{5}{2}$ 【点拨】如图，作$AE⊥ BD$于点E，交PQ于点F，连结EQ并延长交AD于点H，则$∠ AED=∠ AEB=90°. \because$ 四边形ABCD是边长为5的正方形，$\therefore AD=AB=5,∠ BAD=90°,\therefore DE=BE,∠ DAE=45°.\because$ 等腰直角三角形APQ中，$∠ AQP=90°,\therefore AQ=PQ,\therefore ∠ APQ=∠ PAQ=45°.\because ∠ FEB=∠ FQA=90°,∠ PFE=∠ AFQ,\therefore △ PFE∽ △ AFQ,\therefore \frac{FE}{FQ}=\frac{FP}{FA},\therefore \frac{FE}{FP}=\frac{FQ}{FA}.$ 又$\because ∠ QFE=∠ AFP,\therefore △ QFE∽ △ AFP,\therefore ∠ AEH=∠ APQ=45°,\therefore ∠ AHE=90°,\therefore EH// AB,\therefore \frac{BE}{ED}=\frac{AH}{HD},\therefore AH=DH=\frac{1}{2}AD=\frac{5}{2}.\because DQ≥ DH,\therefore DQ≥ \frac{5}{2},\therefore DQ$的最小值为$\frac{5}{2}.$

9. 如图,在正方形ABCD中,E,F分别是边AD,CD上的点,AE=ED,DF=$\frac{1}{4}$DC,连结BE,EF,并延长EF交BC的延长线于点G.
(1)求证:△ABE∽△DEF;

(2)若正方形的边长为4,求BG的长.

9.(1)【证明】$\because$ 四边形ABCD为正方形，$\therefore AD=AB=DC,∠ A=∠ D=90°. \because AE=ED,\therefore \frac{AE}{AB}=\frac{1}{2}. \because DF=\frac{1}{4}DC,\therefore$ 易得$\frac{DF}{DE}=\frac{1}{2}. \therefore \frac{AE}{AB}=\frac{DF}{DE},即\frac{AE}{DF}=\frac{AB}{DE}. \therefore △ ABE∽ △ DEF.$
(2)【解】$\because$ 四边形ABCD为正方形，$\therefore DE// CG.\therefore △ DEF∽ △ CGF.\therefore \frac{DE}{CG}=\frac{DF}{CF}. \because DF=\frac{1}{4}DC,AE=DE,$ 正方形的边长为4，$\therefore DE=2,\frac{DE}{CG}=\frac{DF}{CF}=\frac{1}{3}. \therefore CG=6.\therefore BG=BC+CG=10.$