9. 如图,将一张直角三角形纸片BEC的斜边放在矩形ABCD的BC边上,恰好完全重合,BE,CE分别交AD于点F,G,BC=6,AF:FG:GD=3:2:1,则AB的长为 ()

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

∵ 四边形 ABCD 是矩形,
∴ $AB=CD,AD=BC=6$,
$∠A=∠D=90^{\circ }.\because ∠E=90^{\circ },\therefore ∠EFG+∠EGF=90^{\circ }$,
$\therefore ∠AFB+∠DGC=90^{\circ }.\because ∠AFB+∠ABF=90^{\circ },\therefore ∠ABF=$
$∠DGC,\therefore △ AFB∽ △ DCG,\therefore \frac {AF}{CD}=\frac {AB}{DG}.\because AF:FG:GD=$
$3:2:1,AD=6,\therefore AF=3,DG=1,\therefore AB^{2}=AF· DG=3$,
$\therefore AB=\sqrt {3}$.故选 C.

10. (广元中考)如图,在矩形ABCD中,E是AD边的中点,BE⊥AC,垂足为F,连接DF,下列四个结论:①$△ AEF ∽ △ CAB$;②$\frac{CD}{AD}=\sqrt{2}$;③$DF=DC$;④$CF=2AF$.其中正确的是 ()

A.①②③
B.②③④
C.①③④
D.①②④

如图,过点 D 作 $DM// BE$ 交AC 于点 N.
∵ 四边形 ABCD 是矩
形,
∴ $AD// BC,∠ABC=90^{\circ },AD=BC$.
$\therefore ∠EAC=∠ACB.\because BE⊥ AC$ 于点 F,
$\therefore ∠ABC=∠AFE=90^{\circ },\therefore △ AEF∽$
$△ CAB$,故①正确.
∵ $AD// BC,\therefore △ AEF∽ △ CBF,\therefore \frac {AE}{BC}=$
$\frac {AF}{CF}.\because AE=\frac {1}{2}AD=\frac {1}{2}BC,\therefore \frac {AF}{CF}=\frac {1}{2},\therefore CF=2AF$,故④正
确.
∵ $DE// BM,BE// DM,\therefore$ 四边形 BMDE 是平行四边形,
$\therefore BM=DE=\frac {1}{2}BC,\therefore BM=CM,\therefore CN=NF.\because BE⊥ AC$ 于点
$F,DM// BE,\therefore DN⊥ CF,\therefore DM$ 垂直平分 CF, $\therefore DF=DC$, 故
③正确.设 $AE=a,AB=b$, 则 $AD=2a$, 由 $△ BAE∽ △ ADC$, 得
$\frac {b}{a}=\frac {2a}{b}$, 即 $b=\sqrt {2}a,\therefore \frac {DC}{AD}=\frac {b}{2a}=\frac {\sqrt {2}}{2}$, 故②不正确.正确的结
论是①③④.故选 C.

11. (泰安中考)如图,在矩形ABCD中,点E在DC上,DE=BE,AC与BD相交于点O,BE与AC相交于点F.
(1)若BE平分$∠ CBD$,求证:$BF ⊥ AC$;
(2)找出图中与$△ OBF$相似的三角形,并说明理由;
(3)若$OF=3$,$EF=2$,则$DE$的长度为.

(1) 如图, 在矩形 ABCD 中, $OD=$
$OC,AB// CD,∠BCD=90^{\circ }$,
$\therefore ∠2=∠3=∠4,∠3+∠5=90^{\circ }$.
$\because DE=BE,\therefore ∠1=∠2$.
∵ BE 平分 $∠DBC$,
$\therefore ∠1=∠6,\therefore ∠3=∠6$,
$\therefore ∠6+∠5=90^{\circ },\therefore ∠BFC=90^{\circ }$, 即 $BF⊥ AC$.
(2) 与 $△ OBF$ 相似的三角形有 $△ ECF,△ BAF$, 理由如下:
如图, 在矩形 ABCD 中, $∠4=∠3=∠2.\because ∠1=∠2$,
$\therefore ∠1=∠4$.又
∵ $∠OFB=∠BFA,\therefore △ OBF∽ △ BAF$.
$\because ∠1=∠3,∠OFB=∠EFC,\therefore △ OBF∽ △ ECF$.
(3) $\because △ OBF∽ △ ECF,\therefore \frac {EF}{OF}=\frac {CF}{BF},\therefore \frac {2}{3}=$
$\frac {CF}{BF}$, 即 $3CF=2BF,\therefore 3(CF+OF)=3CF+9=2BF+9$,
$\therefore 3OC=2BF+9,\therefore 3OA=2BF+9$ ①. $\because △ ABF∽ △ BOF$,
$\therefore \frac {OF}{BF}=\frac {BF}{AF},\therefore BF^{2}=OF· AF,\therefore BF^{2}=3(OA+3)$ ②. 联立
①②, 可得 $BF=1+\sqrt {19}$ (负值已舍去), $\therefore DE=BE=2+1+$
$\sqrt {19}=3+\sqrt {19}$.

12. (2024·武汉中考)如图是“赵爽弦图”，它是由四个全等的直角三角形和中间的小正方形MNPQ拼成的一个大正方形ABCD. 直线MP交正方形ABCD的两边于点E,F,记正方形ABCD的面积为$S_1$,正方形MNPQ的面积为$S_2$.若$BE=kAE(k＞1)$,则用含$k$的式子表示$\frac{S_1}{S_2}$的值是.

如图,过点 E 作 $EG⊥ AN$
于点 G,不妨设 $MN=a$,设 $EG=1$,
∵ 四
边形 MNPQ 是正方形, $\therefore ∠PMN=45^{\circ }$,
$\therefore ∠EMG=∠PMN=45^{\circ },\therefore EG=MG=1$.
在 $△ AEG$ 和 $△ ABN$ 中, $∠EAG=∠BAN$,
$∠AGE=∠ANB=90^{\circ },\therefore △ AEG∽ △ ABN,\therefore \frac {AE}{AB}=\frac {EG}{BN}=\frac {AG}{AN}$.
$\because BE=kAE(k＞1),\therefore AB=AE+BE=(k+1)AE,\therefore \frac {AE}{AB}=\frac {1}{k+1}=$
$\frac {AG}{AN},\therefore BN=1+k$. 由题意可知, $△ ABN≌ △ DAM$,
$\therefore BN=AM=1+k,\therefore AG=AM-GM=1+k-1=k,\therefore \frac {AG}{AN}=$
$\frac {AG}{AM+MN}=\frac {k}{k+1+a}=\frac {1}{k+1},\therefore a=k^{2}-1,\therefore AN=AG+GM+MN=k+$
$1+k^{2}-1=k^{2}+k$,
∴ 正方形 ABCD 的面积 $S_{1}=AB^{2}=$
$BN^{2}+AN^{2}=(k+1)^{2}+(k^{2}+k)^{2}=(k+1)^{2}(k^{2}+1)$, 正方形 MNPQ 的
面积 $S_{2}=MN^{2}=a^{2}=(k^{2}-1)^{2}=(k+1)^{2}(k-1)^{2}$,
$\therefore \frac {S_{1}}{S_{2}}=\frac {(k+1)^{2}(k^{2}+1)}{(k+1)^{2}(k-1)^{2}}.\because k＞1,\therefore (k+1)^{2}≠0,\therefore \frac {S_{1}}{S_{2}}=$
$\frac {k^{2}+1}{(k-1)^{2}}$.故答案为 $\frac {k^{2}+1}{(k-1)^{2}}$.

13. (2026·济南期中)在正方形$ABCD$中,$AB=2$,$E$是$BC$的中点,在$BC$延长线上取点$F$使$EF=ED$,过点$F$作$FG⊥ED$交$ED$于点$M$,交$AB$于点$G$,交$CD$于点$N$,以下结论中:
①$\frac{CN}{CF}=\frac{1}{2}$; ②$NM=NC$; ③$\frac{CM}{EG}=\frac{1}{2}$;
④$S_{四边形GBEM}=\frac{\sqrt{5}+1}{2}$.正确的有.

∵ 正方形 ABCD 中, $AB=2$,E 是 BC 的中
点, $\therefore BE=CE=1,CD=AB=2,∠DCE=90^{\circ },\therefore CE:CD=$
$1:2.\because EF=ED,∠DCE=∠FME=90^{\circ },∠DEC=∠MEF$,
$\therefore △ DEC≌ △ FEM,\therefore MF=CD=2,ME=CE=1.\because FG⊥ ED$,
$∠DCF=90^{\circ },\therefore ∠EMF=∠DCF$.又
∵ $∠F=∠F,\therefore △ MEF∽$
$△ CNF,\therefore \frac {CN}{ME}=\frac {CF}{MF},\therefore \frac {CN}{CF}=\frac {ME}{MF}=\frac {1}{2}$,故①正确.
∵ $EF=ED$,
$\therefore EF-CE=ED-EM,\therefore FC=DM$.又
∵ $∠MND=∠FNC$,
$∠DMN=∠FCN,\therefore △ DMN≌ △ FCN,\therefore NM=CN$,故②正确.
$\because BE=EC,ME=EC,\therefore BE=ME$. 在 $Rt△ GBE$ 和
$Rt△ GME$ 中, $BE=ME,GE=GE,\therefore Rt△ GBE≌ Rt△ GME$,
$\therefore ∠BEG=∠MEG.\because ME=EC,\therefore ∠EMC=∠ECM$.又
$\because ∠EMC+∠ECM=∠BEG+∠MEG,\therefore ∠GEB=∠MCE$,
$\therefore MC// GE,\therefore \frac {CM}{EG}=\frac {CF}{EF}.\because EF=DE=\sqrt {EC^{2}+CD^{2}}=\sqrt {5}$,
$\therefore CF=EF-EC=\sqrt {5}-1,\therefore \frac {CM}{EG}=\frac {\sqrt {5}-1}{\sqrt {5}}=\frac {5-\sqrt {5}}{5}$,故③错误.
$\because EF=\sqrt {5},\therefore BF=BE+EF=1+\sqrt {5}.\because CN// BG,\therefore △ GBF∽$
$△ NCF,\therefore \frac {GB}{CN}=\frac {BF}{CF},\therefore \frac {GB}{BF}=\frac {CN}{CF}=\frac {1}{2},\therefore GB=\frac {1}{2}BF=\frac {\sqrt {5}+1}{2}$,
$\therefore S_{△ BGE}=\frac {1}{2}BE×BG=\frac {1}{2}×1×\frac {1+\sqrt {5}}{2}=\frac {1+\sqrt {5}}{4}.\because △ GBE≌$
$△ GME,\therefore S_{四边形GBEM}=2S_{△ BGE}=2×\frac {1+\sqrt {5}}{4}=\frac {1+\sqrt {5}}{2}$,故④正确.
故正确的有①②④.

14.（安徽中考）已知正方形ABCD，点M为边AB的中点.
（1）如图①，点G为线段CM上一点，且∠AGB=90°，延长AG，BG分别与边BC，CD交于点E，F.
求证：①BE=CF；②BE²=BC·CE.
（2）如图②，在边BC上取一点E，满足BE²=BC·CE，连接AE交CM于点G，连接BG并延长交CD于点F，$\frac{FC}{BC}$的值为.

(1) ①
∵ 四边形 ABCD 是正方形, $\therefore AB=BC,∠ABC=$
$∠BCF=90^{\circ },\therefore ∠ABG+∠CBF=90^{\circ }.\because ∠AGB=90^{\circ }$,
$\therefore ∠ABG+∠BAG=90^{\circ },\therefore ∠BAG=∠CBF$.又
∵ $AB=BC$,
$∠ABE=∠BCF=90^{\circ },\therefore △ ABE≌ △ BCF,\therefore BE=CF$.
② $\because ∠AGB=90^{\circ }$, 点 M 为 AB 的中点, $\therefore MG=MA=MB$,
$\therefore ∠GAM=∠AGM$.又
∵ $∠CGE=∠AGM,∠GAM=∠CBG$,
$\therefore ∠CGE=∠CBG$.又 $∠ECG=∠GCB,\therefore △ CGE∽ △ CBG$,
$\therefore \frac {CE}{CG}=\frac {CG}{CB}$, 即 $CG^{2}=BC· CE$. 由 $∠CFG=∠GBM=∠CGF$ 得 $CF=CG$, 由①知 $BE=CF,\therefore BE=CG,\therefore BE^{2}=BC· CE$.
(2) 如图, 延长 AE,
DC 交于点 N,
∵ 四边形 ABCD 是正方形, $\therefore AB// CD,\therefore ∠N=$
$∠EAB$.又
∵ $∠CEN=∠BEA,\therefore △ CEN∽ △ BEA,\therefore \frac {CE}{BE}=\frac {CN}{BA}$,
即 $BE· CN=AB· CE.\because AB=BC,BE^{2}=BC· CE,\therefore CN=BE$.
$\because AB// DN,\therefore \frac {CN}{AM}=\frac {CG}{GM}=\frac {CF}{BM}.\because AM=MB,\therefore FC=CN=$
BE.不妨设正方形的边长为 1, $BE=x$, 由 $BE^{2}=BC· CE$ 可
得 $x^{2}=1· (1-x)$, 解得 $x_{1}=\frac {\sqrt {5}-1}{2},x_{2}=\frac {-\sqrt {5}-1}{2}$ (舍去), $\therefore \frac {FC}{BC}=$
$\frac {BE}{BC}=\frac {\sqrt {5}-1}{2}$.