9. (2026·泰安期末)如图①是装了液体的高脚杯示意图(数据如图),用去一部分液体后如图②所示,此时液面AB=cm.

9. $\dfrac{15}{4}$ 解析:如图,过$O$作$OM⊥ CD$,垂足为$M$,过$O'$作$O'N⊥ AB$,垂足为$N$,$\because CD// AB,\therefore △ CDO∽ △ ABO',\therefore \dfrac{CD}{AB}=\dfrac{OM}{O'N}.\because OM=15-7=8(\mathrm{cm}),O'N=12-7=5(\mathrm{cm}),\therefore \dfrac{6}{AB}=\dfrac{8}{5}$,$\therefore AB=\dfrac{15}{4}.$

10. 在$△ ABC$中，$AB=6$，$AC=5$，点$D$在边$AB$上，且$AD=2$，点$E$在边$AC$上，当$AE=$时，以$A,D,E$为顶点的三角形与$△ ABC$相似.

10. $\dfrac{12}{5}$或$\dfrac{5}{3}$ 解析:设$AE=x$.①若$△ ADE∽ △ ACB$,则$\dfrac{AD}{AC}=\dfrac{AE}{AB}$,即$\dfrac{2}{5}=\dfrac{x}{6}$,解得$x=\dfrac{12}{5}$;②若$△ ADE∽ △ ABC$,则$\dfrac{AD}{AB}=\dfrac{AE}{AC}$,即$\dfrac{2}{6}=\dfrac{x}{5}$,解得$x=\dfrac{5}{3}.$

11.（包头中考）如图，BD是$\odot O$的直径，A是$\odot O$外一点，点C在$\odot O$上，AC与$\odot O$相切于点C，$∠ CAB = 90°$，若$BD = 6$，$AB = 4$，$∠ ABC = ∠ CBD$，则弦BC的长为。

11. $2\sqrt{6}$ 解析:连接$CD,OC$,如图,$\because AC$与$\odot O$相切于点$C$,$\therefore AC⊥ OC. \because ∠ CAB=90°$,$\therefore AC⊥ AB$,$\therefore OC// AB$,$\therefore ∠ ABC=∠ OCB. \because OB=OC$,$\therefore ∠ OCB=∠ CBO$,$\therefore ∠ ABC=∠ CBO. \because BD$是$\odot O$的直径,$\therefore ∠ BCD=90°=∠ CAB$,$\therefore △ ABC∽ △ CBD$,$\therefore \dfrac{AB}{BC}=\dfrac{BC}{BD}$,$\therefore BC^2=AB× BD=4× 6=24$,$\therefore BC=\sqrt{24}=2\sqrt{6}.$

12. 如图,在平面直角坐标系中,矩形OABC的顶点A,B在反比例函数$y=\frac{k}{x}(x＜0)$的图象上,纵坐标分别为1,4,则k的值为.

12. $-2$ 解析:如图,过点$A$作$AD⊥ x$轴于点$D$,过点$B$作$BE⊥ AD$,交$DA$的延长线于点$E$,$\therefore ∠ E=∠ ADO=90°.\because$ 四边形$OABC$为矩形,$\therefore ∠ BAO=90°$,$\therefore ∠ EAB+∠ DAO=90°. \because ∠ EBA+∠ EAB=90°$,$\therefore ∠ DAO=∠ EBA$,$\therefore △ BAE∽ △ AOD$,$\therefore \dfrac{BE}{AD}=\dfrac{AE}{OD}.$ 设$A(k,1)$,$B(\dfrac{k}{4},4)$,则$OD=-k$,$AD=1$,$ED=4$,$\therefore AE=3$,$BE=-\dfrac{3}{4}k$,$\therefore \dfrac{-\dfrac{3}{4}k}{1}=\dfrac{3}{-k}$,解得$k=\pm 2.\because$ 反比例函数图象位于第二象限,$\therefore k=-2.$

13. (2025·内蒙古中考)如图,在菱形ABCD中, $AB=4\sqrt{5}$,对角线BD的长为16,E是AD的中点,F是BD上一点,连接EF.若$BF=3$,则EF的长为.

13. $\sqrt{85}$ 解析:连接$AC$,交$BD$于点$O$,过点$E$作$EG⊥ OD$于点$G$,$\because$ 四边形$ABCD$是菱形,$\therefore AD=AB=4\sqrt{5}$,$BO=OD=\dfrac{1}{2}BD=8$,$AO⊥ BD$,$\therefore AO=\sqrt{AD^2-OD^2}=4$,$EG// AO$,$\therefore △ DEG∽ △ DAO$,$\therefore \dfrac{DE}{AD}=\dfrac{EG}{AO}=\dfrac{DG}{OD}.\because E$是$AD$的中点,$\therefore \dfrac{DE}{AD}=\dfrac{EG}{4}=\dfrac{DG}{8}=\dfrac{1}{2}$,$\therefore EG=2$,$DG=4$,$\therefore FG=BD-BF-DG=16-3-4=9$,$\therefore EF=\sqrt{FG^2+EG^2}=\sqrt{9^2+2^2}=\sqrt{85}.$

14. 如图,在平面直角坐标系中,M,N,C三点的坐标分别为$( \dfrac{1}{2},1 ),(3,1),(3,0)$,点A为线段MN上的一个动点,连接AC,过点A作$AB ⊥ AC$交y轴于点B,当点A从M运动到N时,点B随之运动.设点B的坐标为$(0,b)$,则b的取值范围是.

14. $-\dfrac{5}{4}≤ b≤ 1$ 解析:如图,延长$NM$交$y$轴于点$P$,则$MN⊥ y$轴.连接$CN$,在$△ PAB$和$△ NCA$中,$∠ APB=∠ CNA=90°$,$∠ PAB=∠ NCA=90°-∠ CAN$,$\therefore △ PAB∽ △ NCA$,$\therefore \dfrac{PB}{NA}=\dfrac{PA}{NC}.$ 设$PA=x$,则$NA=PN-PA=3-x$,设$PB=y$,$\therefore \dfrac{y}{3-x}=\dfrac{x}{1}$,$\therefore y=3x-x^2=-(x-\dfrac{3}{2})^2+\dfrac{9}{4}.\because -1＜0$,$\dfrac{1}{2}≤ x≤ 3$,$\therefore$ 当$x=\dfrac{3}{2}$时,$y$有最大值$\dfrac{9}{4}$,此时$b=1-\dfrac{9}{4}=-\dfrac{5}{4}$,当$x=3$时,$y$有最小值0,此时$b=1$,$\therefore b$的取值范围是$-\dfrac{5}{4}≤ b≤ 1.$

三、解答题(共58分)
15.(8分)(黄冈中考)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$EC$的长.

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