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2025年通城学典非常课课通九年级数学下册苏科版江苏专版
注：目前有些书本章节名称可能整理的还不是很完善，但都是按照顺序排列的，请同学们按照顺序仔细查找。练习册 2025年通城学典非常课课通九年级数学下册苏科版江苏专版 答案主要是用来给同学们做完题方便对答案用的，请勿直接抄袭。
8.（2024·兰州）如图，△ABC内接于⊙O，AB为⊙O的直径，D为⊙O上一点，BC = BD，延长BA至点E，使得∠ADE = ∠CBA.
（1）求证：ED是⊙O的切线；
（2）若OB = 4，tan ∠CBA = $\frac{1}{2}$，求ED的长.
（1）求证：ED是⊙O的切线；
（2）若OB = 4，tan ∠CBA = $\frac{1}{2}$，求ED的长.
答案：
(1) 连接$OD$. $\because AB$为$\odot O$的直径，$\therefore \angle BCA=\angle BDA = 90^{\circ}$. 在$Rt\triangle BCA$和$Rt\triangle BDA$中，$\begin{cases}BA = BA,\\BC = BD,\end{cases}$ $\therefore Rt\triangle BCA\cong Rt\triangle BDA$. $\therefore \angle CBA=\angle DBA$. $\because OB = OD$，$\therefore \angle BDO=\angle DBA=\angle CBA$. $\because \angle ADE=\angle CBA$，$\therefore \angle ADE=\angle DBA=\angle BDO$. $\because \angle BDA=\angle BDO+\angle ADO = 90^{\circ}$，$\therefore \angle ODE=\angle ADE+\angle ADO = 90^{\circ}$. $\therefore ED\perp OD$. $\because$ 点$D$在$\odot O$上，$\therefore ED$是$\odot O$的切线
(2) $\because \angle CBA=\angle DBA$，$\therefore \tan\angle DBA=\tan\angle CBA=\frac{1}{2}$. 在$Rt\triangle ABD$中，$\because \angle BDA = 90^{\circ}$，$\therefore \tan\angle DBA=\frac{AD}{BD}=\frac{1}{2}$. $\because \angle ADE=\angle DBA$，$\angle E=\angle E$，$\therefore \triangle EAD\sim\triangle EDB$. $\therefore \frac{ED}{EB}=\frac{EA}{ED}=\frac{AD}{DB}=\frac{1}{2}$. $\therefore ED=\frac{1}{2}EB$，$EA=\frac{1}{2}ED$. $\because OB = 4$，$\therefore AB = 2OB = 8$. $\therefore EB=EA + AB=\frac{1}{2}ED + 8$. $\therefore ED=\frac{1}{2}(\frac{1}{2}ED + 8)$，解得$ED=\frac{16}{3}$. $\therefore ED$的长为$\frac{16}{3}$
(1) 连接$OD$. $\because AB$为$\odot O$的直径，$\therefore \angle BCA=\angle BDA = 90^{\circ}$. 在$Rt\triangle BCA$和$Rt\triangle BDA$中，$\begin{cases}BA = BA,\\BC = BD,\end{cases}$ $\therefore Rt\triangle BCA\cong Rt\triangle BDA$. $\therefore \angle CBA=\angle DBA$. $\because OB = OD$，$\therefore \angle BDO=\angle DBA=\angle CBA$. $\because \angle ADE=\angle CBA$，$\therefore \angle ADE=\angle DBA=\angle BDO$. $\because \angle BDA=\angle BDO+\angle ADO = 90^{\circ}$，$\therefore \angle ODE=\angle ADE+\angle ADO = 90^{\circ}$. $\therefore ED\perp OD$. $\because$ 点$D$在$\odot O$上，$\therefore ED$是$\odot O$的切线
(2) $\because \angle CBA=\angle DBA$，$\therefore \tan\angle DBA=\tan\angle CBA=\frac{1}{2}$. 在$Rt\triangle ABD$中，$\because \angle BDA = 90^{\circ}$，$\therefore \tan\angle DBA=\frac{AD}{BD}=\frac{1}{2}$. $\because \angle ADE=\angle DBA$，$\angle E=\angle E$，$\therefore \triangle EAD\sim\triangle EDB$. $\therefore \frac{ED}{EB}=\frac{EA}{ED}=\frac{AD}{DB}=\frac{1}{2}$. $\therefore ED=\frac{1}{2}EB$，$EA=\frac{1}{2}ED$. $\because OB = 4$，$\therefore AB = 2OB = 8$. $\therefore EB=EA + AB=\frac{1}{2}ED + 8$. $\therefore ED=\frac{1}{2}(\frac{1}{2}ED + 8)$，解得$ED=\frac{16}{3}$. $\therefore ED$的长为$\frac{16}{3}$
9.（2024·邗江一模）如图，在□ABCD中，E是对角线AC上一点，连接BE、DE，且BE = DE.
（1）求证：四边形ABCD是菱形；
（2）若AB = 10，tan ∠BAC = 2，求四边形ABCD的面积.
（1）求证：四边形ABCD是菱形；
（2）若AB = 10，tan ∠BAC = 2，求四边形ABCD的面积.
答案：
(1) 连接$BD$，交$AC$于点$O$. $\because$ 四边形$ABCD$是平行四边形，$\therefore OB = OD$. 在$\triangle BOE$和$\triangle DOE$中，$\begin{cases}OB = OD,\\OE = OE,\\BE = DE,\end{cases}$ $\therefore \triangle BOE\cong\triangle DOE$. $\therefore \angle EOB=\angle EOD = 90^{\circ}$. $\therefore AC\perp BD$. $\therefore$ 四边形$ABCD$是菱形
(2) $\because$ 四边形$ABCD$是菱形，$\therefore AC = 2OA$，$BD = 2OB$. 由
(1)，得$AC\perp BD$，$\therefore \angle AOB = 90^{\circ}$. 在$Rt\triangle AOB$中，$\tan\angle BAO=\tan\angle BAC=\frac{OB}{OA}=2$，$\therefore OB = 2OA$. $\because OB^{2}+OA^{2}=AB^{2}$，$\therefore 4OA^{2}+OA^{2}=10^{2}$. $\therefore OA = 2\sqrt{5}$(负值舍去). $\therefore AC = 2OA = 4\sqrt{5}$. $\therefore BD = 2OB = 8\sqrt{5}$. $\therefore S_{四边形ABCD}=\frac{1}{2}AC\cdot BD=\frac{1}{2}\times4\sqrt{5}\times8\sqrt{5}=80$
(1) 连接$BD$，交$AC$于点$O$. $\because$ 四边形$ABCD$是平行四边形，$\therefore OB = OD$. 在$\triangle BOE$和$\triangle DOE$中，$\begin{cases}OB = OD,\\OE = OE,\\BE = DE,\end{cases}$ $\therefore \triangle BOE\cong\triangle DOE$. $\therefore \angle EOB=\angle EOD = 90^{\circ}$. $\therefore AC\perp BD$. $\therefore$ 四边形$ABCD$是菱形
(2) $\because$ 四边形$ABCD$是菱形，$\therefore AC = 2OA$，$BD = 2OB$. 由
(1)，得$AC\perp BD$，$\therefore \angle AOB = 90^{\circ}$. 在$Rt\triangle AOB$中，$\tan\angle BAO=\tan\angle BAC=\frac{OB}{OA}=2$，$\therefore OB = 2OA$. $\because OB^{2}+OA^{2}=AB^{2}$，$\therefore 4OA^{2}+OA^{2}=10^{2}$. $\therefore OA = 2\sqrt{5}$(负值舍去). $\therefore AC = 2OA = 4\sqrt{5}$. $\therefore BD = 2OB = 8\sqrt{5}$. $\therefore S_{四边形ABCD}=\frac{1}{2}AC\cdot BD=\frac{1}{2}\times4\sqrt{5}\times8\sqrt{5}=80$
10.（2023·杭州）第二十四届国际数学家大会会徽的设计基础是1700多年前我国古代数学家赵爽的“弦图”.如图，在由四个全等的直角三角形（Rt △DAE、Rt △ABF、Rt △BCG、Rt△CDH）和中间一个小正方形EFGH拼成的大正方形ABCD中，∠ABF＞∠BAF，连接BE.设∠BAF = α，∠BEF = β，若正方形EFGH与正方形ABCD的面积之比为1∶n，tan α = tan²β，则n的值为（ ）
A. 5
B. 4
C. 3
D. 2
A. 5
B. 4
C. 3
D. 2
答案：
C 解析：设$AF = a$，$BF = b$，则易得$AE = BF = b$. $\therefore EF=AF - AE=a - b$. 由题意，得$S_{正方形ABCD}=AD^{2}=b^{2}+a^{2}$，$S_{正方形EFGH}=EF^{2}=(a - b)^{2}$. $\because \angle AFB = 90^{\circ}$，$\therefore \tan\alpha=\tan\angle BAF=\frac{BF}{AF}=\frac{b}{a}$. $\because \angle EFB = 90^{\circ}$，$\therefore \tan\beta=\tan\angle BEF=\frac{BF}{EF}=\frac{b}{a - b}$. $\because \tan\alpha=\tan^{2}\beta$，$\therefore \frac{b}{a}=(\frac{b}{a - b})^{2}$. $\therefore (a - b)^{2}=ab$. $\therefore b^{2}+a^{2}=3ab$. $\therefore S_{正方形EFGH}:S_{正方形ABCD}=ab:3ab = 1:3$. $\because S_{正方形EFGH}:S_{正方形ABCD}=1:n$，$\therefore 1:n = 1:3$. $\therefore n = 3$.
11. 如图，在△ABC中，若∠A = 45°，AC² - BC² = $\frac{\sqrt{5}}{5}$AB²，则tan C = _______.
答案：
$\sqrt{5}$ 解析：过点$B$作$BD\perp AC$于点$D$. $\therefore \angle ADB=\angle CDB = 90^{\circ}$. $\therefore \angle ABD = 90^{\circ}-\angle A = 90^{\circ}-45^{\circ}=45^{\circ}$. $\therefore \angle A=\angle ABD$. $\therefore AD = BD$. 设$BD = AD = x$，$CD = y(x\gt0,y\gt0)$. 在$Rt\triangle BCD$中，由勾股定理，得$BC^{2}=BD^{2}+CD^{2}=x^{2}+y^{2}$. 在$Rt\triangle ABD$中，由勾股定理，得$AB^{2}=AD^{2}+BD^{2}=x^{2}+x^{2}=2x^{2}$. $\because AC^{2}-BC^{2}=\frac{\sqrt{5}}{5}AB^{2}$，$\therefore (x + y)^{2}-(x^{2}+y^{2})=\frac{\sqrt{5}}{5}\times2x^{2}$. 化简，得$y=\frac{\sqrt{5}}{5}x$. 在$Rt\triangle BCD$中，$\tan C=\frac{BD}{CD}=\frac{x}{y}=\frac{x}{\frac{\sqrt{5}}{5}x}=\sqrt{5}$.
12.（2023·成都）如图，以△ABC的边AC为直径作⊙O，交边BC于点D，过点C作CE//AB，交⊙O于点E，连接AD、DE，∠B = ∠ADE.
（1）求证：AC = BC；
（2）若tan B = 2，CD = 3，求AB和DE的长.
（1）求证：AC = BC；
（2）若tan B = 2，CD = 3，求AB和DE的长.
答案：
(1) $\because CE// AB$，$\therefore \angle BAC=\angle ACE$. $\because \overset{\frown}{AE}=\overset{\frown}{AE}$，$\therefore \angle BAC=\angle ACE=\angle ADE$. $\because \angle B=\angle ADE$，$\therefore \angle B=\angle BAC$. $\therefore AC = BC$
(2) 设$BD = x$. $\because AC$是$\odot O$的直径，$\therefore \angle ADC=\angle ADB = 90^{\circ}$. $\because \tan B = 2$，$\therefore \frac{AD}{BD}=2$，即$AD = 2x$. 根据
(1)中的结论，可得$AC = BC = BD + CD=x + 3$. 在$Rt\triangle ADC$中，根据勾股定理，可得$AD^{2}+DC^{2}=AC^{2}$，即$(2x)^{2}+3^{2}=(x + 3)^{2}$，解得$x_{1}=2$，$x_{2}=0$(不合题意，舍去). $\therefore BD = 2$，$AD = 4$. 在$Rt\triangle ABD$中，根据勾股定理，可得$AB=\sqrt{AD^{2}+BD^{2}}=2\sqrt{5}$. 如图，过点$E$作$EF\perp DC$，交$DC$的延长线于点$F$. $\because CE// AB$，$\therefore \angle ECF=\angle B$. $\because EF\perp CF$，$\therefore \tan\angle ECF=\tan B = 2$，即$\frac{EF}{CF}=2$. $\because \angle B+\angle BAD = 90^{\circ}$，$\angle ADE+\angle EDF = 90^{\circ}$，$\angle B=\angle ADE$，$\therefore \angle BAD=\angle EDF$. $\therefore \angle DEF = 90^{\circ}-\angle EDF = 90^{\circ}-\angle BAD=\angle B$. $\therefore \tan\angle DEF=\tan B=\frac{DF}{EF}=2$. 设$CF = a$，则$DF = CF + CD=a + 3$，$EF = 2a$. $\therefore \frac{a + 3}{2a}=2$，解得$a = 1$. 经检验，$a = 1$是原分式方程的解，且符合题意. $\therefore EF = 2$，$DF = 4$. 在$Rt\triangle DEF$中，根据勾股定理，可得$DE=\sqrt{DF^{2}+EF^{2}}=2\sqrt{5}$
(1) $\because CE// AB$，$\therefore \angle BAC=\angle ACE$. $\because \overset{\frown}{AE}=\overset{\frown}{AE}$，$\therefore \angle BAC=\angle ACE=\angle ADE$. $\because \angle B=\angle ADE$，$\therefore \angle B=\angle BAC$. $\therefore AC = BC$
(2) 设$BD = x$. $\because AC$是$\odot O$的直径，$\therefore \angle ADC=\angle ADB = 90^{\circ}$. $\because \tan B = 2$，$\therefore \frac{AD}{BD}=2$，即$AD = 2x$. 根据
(1)中的结论，可得$AC = BC = BD + CD=x + 3$. 在$Rt\triangle ADC$中，根据勾股定理，可得$AD^{2}+DC^{2}=AC^{2}$，即$(2x)^{2}+3^{2}=(x + 3)^{2}$，解得$x_{1}=2$，$x_{2}=0$(不合题意，舍去). $\therefore BD = 2$，$AD = 4$. 在$Rt\triangle ABD$中，根据勾股定理，可得$AB=\sqrt{AD^{2}+BD^{2}}=2\sqrt{5}$. 如图，过点$E$作$EF\perp DC$，交$DC$的延长线于点$F$. $\because CE// AB$，$\therefore \angle ECF=\angle B$. $\because EF\perp CF$，$\therefore \tan\angle ECF=\tan B = 2$，即$\frac{EF}{CF}=2$. $\because \angle B+\angle BAD = 90^{\circ}$，$\angle ADE+\angle EDF = 90^{\circ}$，$\angle B=\angle ADE$，$\therefore \angle BAD=\angle EDF$. $\therefore \angle DEF = 90^{\circ}-\angle EDF = 90^{\circ}-\angle BAD=\angle B$. $\therefore \tan\angle DEF=\tan B=\frac{DF}{EF}=2$. 设$CF = a$，则$DF = CF + CD=a + 3$，$EF = 2a$. $\therefore \frac{a + 3}{2a}=2$，解得$a = 1$. 经检验，$a = 1$是原分式方程的解，且符合题意. $\therefore EF = 2$，$DF = 4$. 在$Rt\triangle DEF$中，根据勾股定理，可得$DE=\sqrt{DF^{2}+EF^{2}}=2\sqrt{5}$
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