《2026年通城学典全国中考试题分类精粹中考数学》

18. (2025·绥化)如图，$\angle APO=\angle BPO$，$PA$与$\odot O$相切于点$M$，连接$OM$，$OP$与$\odot O$相交于点$C$，过点$C$作$CD\perp OM$，垂足为$E$，交$\odot O$于点$D$，连接$PD$交$OM$于点$F$。
(1) 求证：$PB$是$\odot O$的切线；
(2) 当$PC = 6$，$PM=\frac{5}{4}CD$时，求线段$MF$的长。

答案： 18.
(1)过点$O$作$ON\perp PB$于点$N$。$\because ON\perp PB$，$\therefore\angle PNO = 90^{\circ}$。$\because PA$与$\odot O$相切于点$M$，$\therefore OM\perp PA$。$\therefore\angle PMO = \angle PNO = 90^{\circ}$。$\because\angle APO = \angle BPO$，$PO = PO$，$\therefore\triangle PMO\cong\triangle PNO$。$\therefore OM = ON$。$\because OM$为$\odot O$的半径，$\therefore ON$为$\odot O$的半径。$\because ON\perp PB$，$\therefore PB$是$\odot O$的切线。
(2)$\because CD\perp OM$，$OM$为半径，$\therefore CE = DE=\frac{1}{2}CD$。$\because PM=\frac{5}{4}CD$，
$\therefore\frac{CD}{PM}=\frac{4}{5}$。$\therefore\frac{CE}{PM}=\frac{2}{5}$。$\because\angle OMP = 90^{\circ}$，$\angle OEC = 90^{\circ}$，
$\therefore CD// PM$。$\therefore\triangle OMP\sim\triangle OEC$。$\therefore\frac{CE}{PM}=\frac{OC}{OP}$。$\because PC = 6$，
$\therefore\frac{2}{5}=\frac{OC}{OC + 6}$。$\therefore OC = 4$。$\therefore OC = OM = 4$。在$Rt\triangle MOP$中，$PM=\sqrt{OP^{2}-OM^{2}}=\sqrt{(6 + 4)^{2}-4^{2}}=2\sqrt{21}$，$\therefore CE = DE=\frac{4\sqrt{21}}{5}$，$OE=\sqrt{OC^{2}-CE^{2}}=\sqrt{4^{2}-(\frac{4\sqrt{21}}{5})^{2}}=\frac{8}{5}$。
$\because\angle FMP=\angle FED$，$\angle MFP=\angle EFD$，$\therefore\triangle MFP\sim\triangle EFD$。$\therefore\frac{MF}{EF}=\frac{MP}{ED}$。设$MF = x$，则$EF = 4 - x-\frac{8}{5}=\frac{12}{5}-x$，$\therefore\frac{x}{\frac{12}{5}-x}=\frac{2\sqrt{21}}{\frac{4\sqrt{21}}{5}}$，解得$x=\frac{12}{7}$。$\therefore MF=\frac{12}{7}$。

19. (2025·上海)已知四边形$ABCD$是平行四边形，$E$，$F$分别是边$BC$，$CD$上的点。
(1) 若$E$是边$BC$的中点：
① 如图①，若$AE = EF$，求证：$\angle BAE=\angle EFC$；
② 如图②，若$CF = DF$，连接$BF$交$AE$于点$G$，求$S_{\triangle BEG}:S_{\triangle AEF}$的值。
(2) 如图③，若$AB = 3$，$AD = 5$，$CF = 1$，$\angle AEB=\angle AFE=\angle EFC$，求$AF$的长。

答案： 19.
(1)①延长$FE$，$AB$交于点$H$。$\because$四边形$ABCD$是平行四边形，$\therefore AB// CD$。$\therefore\angle EBH = \angle ECF$，$\angle H = \angle EFC$。$\because E$是边$BC$的中点，$\therefore BE = CE$。$\therefore\triangle BEH\cong\triangle CEF$。$\therefore EH = EF$。$\because AE = EF$，$\therefore AE = EH$。$\therefore\angle H = \angle BAE$。$\therefore\angle BAE = \angle CFE$。②延长$BF$，$AD$交于点$M$。$\because$四边形$ABCD$是平行四边形，$\therefore AD// BC$，$AD = BC$。
$\therefore\triangle BEG\sim\triangle MAG$，$\triangle BCF\sim\triangle MDF$。$\therefore\frac{BE}{MA}=\frac{GE}{GA}=\frac{BG}{MG}$，$\frac{BC}{MD}=\frac{BF}{MF}=\frac{CF}{DF}=1$。$\therefore BF = MF$，$BC = DM$。$\because E$是边$BC$的中点，$\therefore BC = 2CE = 2BE$。设$CE = BE = m$，则$BC = DM = 2m$，$\therefore AM = AD + DM = 4m$。$\therefore\frac{BE}{AM}=\frac{m}{4m}=\frac{1}{4}$。$\because\frac{BF}{MF}=1$，$\therefore\frac{BG}{GF}=\frac{2}{3}$。$\therefore\frac{S_{\triangle BGE}}{S_{\triangle BGA}}=\frac{S_{\triangle GFE}}{S_{\triangle GFA}}=\frac{GE}{AG}=\frac{1}{4}$，$\frac{S_{\triangle ABG}}{S_{\triangle AFG}}=\frac{BG}{FG}=\frac{2}{3}$。设$S_{\triangle ABG}=4n$，则$S_{\triangle BGE}=n$，$S_{\triangle AFG}=6n$，$\therefore S_{\triangle EGF}=\frac{3}{2}n$。$\therefore\frac{S_{\triangle BGE}}{S_{\triangle AEF}}=\frac{S_{\triangle BGE}}{S_{\triangle AFG}+S_{\triangle EGF}}=\frac{n}{6n+\frac{3}{2}n}=\frac{2}{15}$。
(2)延长$AD$，$EF$交于点$M$。四边形$ABCD$是平行四边形，$\therefore AD// BC$，$CD = AB = 3$。$\therefore\angle AEB = \angle EAD$。$\because\angle AEB = \angle AFE = \angle EFC$，$\therefore\angle EFA = \angle EAD$。又$\because\angle AEF = \angle MEA$，
$\therefore\triangle AEF\sim\triangle MEA$。$\because\angle AEB + \angle AEF + \angle FEC = \angle EFC + \angle FCE + \angle FEC = 180^{\circ}$，$\angle AEB = \angle EFC$，$\therefore\angle AEF = \angle FCE$。$\therefore\triangle AEF\sim\triangle ECF$。$\because AD// BC$，
$\therefore\triangle ECF\sim\triangle MDF$。$\therefore\frac{EC}{MD}=\frac{EF}{MF}=\frac{CF}{DF}$。$\because CF = 1$，
$\therefore DF = CD - CF = 2$。设$CE = s$，$FE = t$。$\because\triangle AEF\sim\triangle ECF$，$\therefore\frac{CF}{EF}=\frac{CE}{EA}=\frac{EF}{AF}$，即$\frac{1}{t}=\frac{s}{AE}=\frac{t}{AF}$。$\therefore AE = st$，$AF = t^{2}$。$\because\frac{EC}{DM}=\frac{EF}{FM}=\frac{CF}{DF}$，即$\frac{s}{DM}=\frac{t}{FM}=\frac{1}{2}$，$\therefore DM = 2s$，$FM = 2t$。$\therefore AM = AD + DM = 5 + 2s$。$\because\triangle AEF\sim\triangle MEA$，$\therefore\frac{EF}{EA}=\frac{AE}{MA}=\frac{AF}{MA}$，即$\frac{t}{st}=\frac{st}{t + 2t}=\frac{t^{2}}{5 + 2s}$。
$\therefore\begin{cases}s=\sqrt{3}\\t^{2}=\frac{5\sqrt{3}+6}{3}\end{cases}$或$\begin{cases}s=-\sqrt{3}\\t^{2}=\frac{6 - 5\sqrt{3}}{3}\end{cases}$（舍去）。$\therefore AF=\frac{5\sqrt{3}+6}{3}$。