《2026年山东省中考试卷精选九年级数学》

22. (本小题满分 10 分)如图，$\triangle ABC$内接于$\odot O$，$\angle ABC=2\angle C$，点 $D$ 在线段 $CB$ 的延长线上，且 $BD=AB$，连接 $AD$.

(1)求证：$AD$是$\odot O$的切线；
(2)当$AB=5$，$AC=8$时，求 $BC$ 的长及$\odot O$的半径.

答案：
22.[解析]本题考查了相似三角形的判定与性质:在判定两个三角形相似时,应注意利用图形中已有的公共角、公共边等隐含条件,以充分发挥基本图形的作用.灵活运用相似三角形的性质计算相应线段的长或表示线段之间的关系是解决问题的关键.也考查了圆周角定理和切线的判定与性质.
(1)证明:如图,作直径AE交$\odot O$于点E,连接BE, $\because BD = AB$,$\therefore \angle D = \angle BAD$, $\therefore \angle ABC = \angle D + \angle BAD = 2\angle BAD$. $\because \angle ABC = 2\angle C$,$\therefore \angle C = \angle BAD$. $\because \angle E = \angle C$,$\therefore \angle E = \angle BAD$. $\because AE$为直径,$\therefore \angle ABE = 90°$, $\therefore \angle E + \angle BAE = 90°$, $\therefore \angle BAD + \angle BAE = 90°$, 即$\angle DAE = 90°$,$\therefore AE \perp AD$. $\because AE$为直径,$\therefore AD$是$\odot O$的切线.
(2)解:$\because \angle D = \angle C$, $\therefore AD = AC = 8$. $\because \angle BAD = \angle C$,$\angle ADB = \angle CDA$, $\therefore \triangle DAB \sim \triangle DCA$, $\therefore DB:DA = DA:DC$, 即$5:8 = 8:DC$,解得$DC = \frac{64}{5}$, $\therefore BC = \frac{64}{5} - 5 = \frac{39}{5}$. 如图,作$AH \perp BC$于点H. $\because AD = AC$,$AH \perp DC$, $\therefore CH = \frac{1}{2}DC = \frac{32}{5}$. 在$Rt \triangle ACH$中,$AC = 8$,$CH = \frac{32}{5}$, $\therefore AH = \sqrt{8^2 - (\frac{32}{5})^2} = \frac{24}{5}$. $\because AE$为直径,$\therefore \angle ABE = 90°$. $\because \angle E = \angle C$,$\angle ABE = \angle AHC$, $\therefore \triangle ABE \sim \triangle AHC$, $\therefore AE:AC = AB:AH$, 即$AE:8 = 5:\frac{24}{5}$,解得$AE = \frac{25}{3}$, $\therefore \odot O$的半径为$\frac{25}{6}$.

23. (本小题满分 11 分)[问题呈现]
如图(1)，已知 $P$ 是正方形$A_{1}A_{2}A_{3}A_{4}$外一点，且满足$\angle PA_{1}A_{2}+\angle PA_{3}A_{2}=180^{\circ}$，探究$PA_{1}$，$PA_{2}$，$PA_{3}$三条线段的数量关系.
小颖通过观察、分析、思考，形成了如下思路：
思路一：如图(2)，构造$\triangle QA_{3}A_{2}$与$\triangle PA_{1}A_{2}$全等，从而得出$PA_{1}+PA_{3}$与$PA_{2}$的数量关系；
思路二：如图(3)，构造$\triangle MA_{1}A_{2}$与$\triangle NA_{3}A_{2}$全等，从而得出$PA_{1}+PA_{3}$与$PA_{2}$的数量关系.

[类比探究]
(1)请参考小颖的思路，直接写出$PA_{1}+PA_{3}$与$PA_{2}$的数量关系$PA_1 + PA_3 = \sqrt{2}PA_2$；
[拓展延伸]
(2)如图(4)，若 $P$ 是正五边形$A_{1}A_{2}A_{3}A_{4}A_{5}$外一点，且满足$\angle PA_{1}A_{2}+\angle PA_{3}A_{2}=180^{\circ}$，$PA_{1}=11$，$PA_{3}=49$，求$PA_{2}$的长度(结果精确到 0.1，参考数据：$\sin 54^{\circ} \approx 0.81$，$\sin 72^{\circ} \approx 0.95$，$\cos 54^{\circ} \approx 0.59$，$\cos 72^{\circ} \approx 0.31$)；
(3)如图(5)，若 $P$ 是正十边形$A_{1}A_{2}·s A_{10}$外一点，且满足$\angle PA_{1}A_{2}+\angle PA_{3}A_{2}=180^{\circ}$，则$PA_{1}$，$PA_{2}$，$PA_{3}$三条线段的数量关系为 $PA_1 + PA_3 = 2PA_2 \sin 72°$(结果用含有锐角三角函数的式子表示).

答案：
23.[解析]本题考查了正方形的性质、全等三角形的性质、解直角三角形、多边形的内角和问题. 解:
(1)$PA_1 + PA_3 = \sqrt{2}PA_2$提示: 如图
(1),在射线$PA_3$上截取$A_3Q = PA_1$,连接$A_2Q$. $\because \angle PA_2A_1 + \angle PA_3A_2 = 180°$, $\angle QA_3A_2 + \angle PA_3A_2 = 180°$, $\therefore \angle PA_2A_1 = \angle A_2A_3Q$. 又四边形$A_1A_2A_3A_4$是正方形, $\therefore A_2A_1 = A_2A_3$, $\therefore \triangle QA_3A_2 \cong \triangle PA_1A_2$, $\therefore \angle A_1A_2P = \angle A_2A_3Q$,$A_2P = A_2Q$. 又四边形$A_1A_2A_3A_4$是正方形, $\therefore \angle A_1A_2A_3 = 90°$,$\therefore \angle PA_2Q = 90°$, $\therefore \triangle A_2PQ$是等腰直角三角形, $\therefore PQ = PA_3 + A_3Q = PA_3 + PA_1 = \sqrt{2}PA_2$.
(2)易知正五边形的一个内角度数为$\frac{(5 - 2) × 180°}{5} = 108°$. 如图
(2),在射线$PA_3$上截取$A_3Q = PA_1$,连接$A_2Q$,过点$A_2$作$AT \perp PQ$于点T, 同
(1)可得$\triangle QA_3A_2 \cong \triangle PA_1A_2$, $\therefore \angle PA_2Q = \angle A_1A_2A_3 = 108°$,$A_2P = A_2Q$. $\therefore \angle PA_2T = \frac{1}{2}\angle PA_2Q = 54°$. $\because PA_1 = 11$,$PA_3 = 49$, $\therefore PQ = PA_3 + A_3Q = PA_3 + PA_1 = 60$,$\therefore PT = \frac{1}{2}PQ = 30$, $\therefore PA_2 = \frac{PT}{\sin 54°} \approx \frac{30}{0.81} \approx 37.0$.
(3)$PA_1 + PA_3 = 2PA_2 \sin 72°$提示:如图
(3),在射线$PA_3$上截取$A_3Q = PA_1$,连接$A_2Q$,过点$A_2$作$A_2T \perp PQ$于点T, 同
(2)可得$\angle PA_2Q = \angle A_1A_2A_3 = (10 - 2) × 180° ÷ 10 = 144°$, $\therefore \angle PA_2T = \frac{1}{2}\angle PA_2Q = 72°$, $\therefore A_2T = PA_2 · \cos 72° = \frac{PT}{\tan 72°}$. $\because 2PT = PA_3 + PA_1$, $\therefore PA_2 · \cos 72° = \frac{PA_3 + PA_1}{2\tan 72°}$. $\therefore PA_1 + PA_3 = 2PA_2 · \sin 72°$.