《2025年全练单元卷九年级数学下册人教版》

26. (10分)如图，反比例函数$y = \frac{m}{x}(x\gt0)$的图象经过线段$OA$的端点$A$，$O$为原点，作$AB\perp x$轴于点$B$，点$B$的坐标为$(2,0)$，$\tan\angle AOB = \frac{3}{2}$.
(1)求$m$的值；
(2)线段$AB$为边向右作矩形$ABCD$，反比例函数$y = \frac{m}{x}(x\gt0)$的图象恰好经过$DC$的中点$E$，求直线$AE$的函数解析式；
(3)在(2)的条件下，若直线$AE$与$x$轴交于点$M$，与$y$轴交于点$N$，问线段$AN$与线段$ME$的大小关系如何？请说明理由.
　　　

答案：
解：
(1)$\because B(2,0),\therefore OB = 2$，$\because \tan\angle AOB=\frac{AB}{OB}=\frac{3}{2},\therefore AB = 3,\therefore A(2,3)$，$\because$反比例函数$y=\frac{m}{x}(x ＞ 0)$的图象经过线段$OA$的端点$A$，$\therefore m = 2×3 = 6$；
(2)$\because A(2,3),B(2,0)$，$\therefore$线段$AB$的中点纵坐标为$\frac{3}{2}$，$\because$四边形$ABCD$为矩形，$\therefore AB = CD$，$\therefore$线段$CD$的中点$E$的纵坐标为$\frac{3}{2}$，由
(1)可知反比例函数解析式为$y=\frac{6}{x}$，当$y=\frac{3}{2}$时，可得$\frac{3}{2}=\frac{6}{x}$，解得$x = 4,\therefore E(4,\frac{3}{2})$，设直线$AE$的函数解析式为$y = kx + b$，把$A,E$坐标代入可得$\begin{cases}2k + b = 3\\4k + b=\frac{3}{2}\end{cases}$，解得$\begin{cases}k = -\frac{3}{4}\\b=\frac{9}{2}\end{cases}$，$\therefore$直线$AE$的函数解析式为$y = -\frac{3}{4}x+\frac{9}{2}$；
(3)相等.理由如下：连接$OE$，延长$DA$交$y$轴于点$F$，

则$AF\perp ON$，且$AF = 2$，对于$y = -\frac{3}{4}x+\frac{9}{2}$，当$y = 0$时，$x = 6$，当$x = 0$时，$y=\frac{9}{2}$，$\therefore M(6,0),N(0,\frac{9}{2}),\therefore OM = 6,ON=\frac{9}{2}$，$\because S_{\triangle EOM}=\frac{1}{2}\cdot OM\cdot EC=\frac{1}{2}×6×\frac{3}{2}=\frac{9}{2}$，$S_{\triangle AON}=\frac{1}{2}ON\cdot AF=\frac{1}{2}×\frac{9}{2}×2=\frac{9}{2}$，$\therefore S_{\triangle EOM}=S_{\triangle AON}$，$\because$在$\triangle MON$中，$AN$和$ME$边上的高相等，$\therefore AN = ME$.

27. (10分)如图，在矩形$ABCD$中，$E$为$CD$的中点，$F$为$BE$上的一点，连接$CF$并延长交$AB$于点$M$，$MN\perp CM$交射线$AD$于点$N$.
(1)当$F$为$BE$中点时，求证：$AM = CE$；
(2)若$\frac{AB}{BC}=\frac{EF}{BF}=2$，求$\frac{AN}{ND}$的值；
(3)若$\frac{AB}{BC}=\frac{EF}{BF}=n$，当$n$为何值时，$MN// BE$？

答案：
(1)证明：当$F$为$BE$中点时，则有$BF = EF$. $\because$四边形$ABCD$是矩形，$\therefore AB = DC,AB// DC$，$\therefore \angle MBF=\angle CEF,\angle BMF=\angle ECF$. 在$\triangle BMF$和$\triangle ECF$中，$\begin{cases}\angle MBF=\angle CEF\\\angle BMF=\angle ECF\\BF = EF\end{cases}$，$\therefore \triangle BMF\cong\triangle ECF(AAS),\therefore BM = EC$. $\because E$为$CD$的中点，$\therefore EC=\frac{1}{2}DC,\therefore BM = EC=\frac{1}{2}DC=\frac{1}{2}AB$，$\therefore AM = BM = EC$；
(2)解：设$MB = a$，$\because$四边形$ABCD$是矩形，$\therefore AD = BC,AB = DC,\angle A=\angle ABC=\angle BCD = 90^{\circ},AB// DC$，$\therefore \triangle ECF\sim\triangle BMF,\therefore \frac{EC}{BM}=\frac{EF}{BF}=2,\therefore EC = 2a$，$\therefore AB = CD = 2CE = 4a,AM = AB - MB = 3a$. $\because \frac{AB}{BC}=2,\therefore BC = AD = 2a$. $\because MN\perp MC,\therefore \angle CMN = 90^{\circ}$，$\therefore \angle AMN+\angle BMC = 90^{\circ}$. $\because \angle A = 90^{\circ},\therefore \angle ANM+\angle AMN = 90^{\circ}$，$\therefore \angle BMC=\angle ANM$，$\therefore \triangle AMN\sim\triangle BCM$，$\therefore \frac{AN}{BM}=\frac{AM}{BC},\therefore \frac{AN}{a}=\frac{3a}{2a}$，$\therefore AN=\frac{3}{2}a,ND = AD - AN = 2a-\frac{3}{2}a=\frac{1}{2}a,\therefore \frac{AN}{ND}=3$；
(3)解：当$\frac{AB}{BC}=\frac{EF}{BF}=n$时，设$MB = a$. $\because AB// CD,\therefore \triangle MFB\sim\triangle CFE$，$\therefore \frac{MB}{EC}=\frac{BF}{EF}=\frac{1}{n}$，即$\frac{a}{EC}=\frac{1}{n}$，解得$EC = an.\therefore AB = 2an$. 又$\because \frac{AB}{BC}=n,\therefore \frac{2an}{BC}=n,\therefore BC = 2a$. $\because MN// BE,MN\perp MC$，$\therefore \angle EFC=\angle NMC = 90^{\circ},\therefore \angle FCB+\angle FBC = 90^{\circ}$. $\because \angle MBC = 90^{\circ},\therefore \angle BMC+\angle FCB = 90^{\circ}$，$\therefore \angle BMC=\angle FBC$. $\because \angle MBC=\angle BCE = 90^{\circ},\therefore \triangle MBC\sim\triangle BCE$，$\therefore \frac{MB}{BC}=\frac{BC}{CE},\therefore \frac{a}{2a}=\frac{2a}{na},\therefore n = 4$.