《2026年通城学典全国中考试题分类精粹中考数学》

20. (2025·福建)如图，四边形$ABCD$内接于$\odot O$，$AD$，$BC$的延长线相交于点$E$，$AC$，$BD$相交于点$F$。$G$是$AB$上一点，$GD$交$AC$于点$H$，且$AB = AC$，$BG = DG$。
(1) 求证：$\angle ABC=\angle DBE+\angle E$；
(2) 求证：$AH^{2}=HF· HC$；
(3) 若$\tan\angle ABC=\sqrt{5}$，$AD = 2DE$，$CD=\sqrt{6}$，求$\triangle AGH$的周长。

答案： 20.
(1)$\because AB = AC$，$\therefore\angle ABC = \angle ACB$。$\because\overset{\frown}{AB}=\overset{\frown}{AB}$，
$\therefore\angle ACB = \angle ADB$。$\therefore\angle ABC = \angle ADB$。$\because\overset{\frown}{AD}=\overset{\frown}{AD}$，$\therefore\angle ABD = \angle ACD$。$\because BG = DG$，$\therefore\angle BDG = \angle ABD = \angle ACD$。又$\because\angle DHF = \angle CHD$，$\therefore\triangle HDF\sim\triangle HCD$。$\therefore\frac{HF}{HD}=\frac{HD}{HC}$。$\therefore HD^{2}=HF· HC$。由
(1)知，
$\angle ABC = \angle DBE + \angle E$，又$\because\angle ABC = \angle DBE + \angle ABD$，
$\therefore\angle ABD = \angle E$。$\therefore\angle BDG = \angle E$。$\because\angle ADB = \angle ADG + \angle BDG$，$\angle ADC = \angle BDC + \angle E$，$\therefore\angle ADG = \angle DAC$。$\therefore AH = DH$。$\therefore AH^{2}=HF· HC$。
(3)由
(2)知，$AH = DH$，
$\therefore\triangle AGH$的周长为$AG + GH + AH = AG + GH + DH = AG + DG = AG + BG = AB$。设$DE = m$，则$AD = 2DE = 2m$，
$AE = AD + DE = 3m$。由
(2)可知，$\angle ACD = \angle E$，又$\because\angle CAD = \angle EAC$，$\therefore\triangle ACD\sim\triangle AEC$。$\therefore\frac{AC}{AE}=\frac{AD}{AC}=\frac{CD}{EC}$。$\therefore AC^{2}=AD· AE = 6m^{2}$。$\therefore AC=\sqrt{6}m$。又$\because CD=\sqrt{6}$，$\therefore\frac{\sqrt{6}m}{3m}=\frac{\sqrt{6}}{EC}$。$\therefore EC = 3$。过点$C$作$CP\perp AE$，垂足为$P$，则$\angle CPD = \angle CPE = 90^{\circ}$。$\because$四边形$ABCD$是圆内接四边形，$\therefore\angle ADC + \angle ABC = 180^{\circ}$。又$\because\angle ADC + \angle CDP = 180^{\circ}$，$\therefore\angle CDP = \angle ABC$。$\therefore\tan\angle CDP=\tan\angle ABC=\sqrt{5}$。
$\therefore$在$Rt\triangle DCP$中，$\frac{PC}{PD}=\sqrt{5}$，即$PC=\sqrt{5}PD$。$\therefore CD=\sqrt{PD^{2}+PC^{2}}=\sqrt{6}PD$。$\therefore\sqrt{6}PD=\sqrt{6}$，$\therefore PD = 1$。$\therefore PE = DE - DP = m - 1$。在$Rt\triangle CPE$中，$PE^{2}+PC^{2}=CE^{2}$，$\therefore(m - 1)^{2}+(\sqrt{5})^{2}=3^{2}$，解得$m = 3$或$m = -1$（舍去）。$\therefore AB = AC = 3\sqrt{6}$。$\therefore\triangle AGH$的周长为$3\sqrt{6}$。

21. (2025·成都)如图，在$□ ABCD$中，点$E$在$BC$边上，点$B$关于直线$AE$的对称点$F$落在$□ ABCD$内，射线$AF$交射线$DC$于点$G$，交射线$BC$于点$P$，射线$EF$交$CD$边于点$Q$。
【特例感知】
(1) 如图①，当$CE = BE$时，点$P$在$BC$的延长线上，求证：$\triangle EFP\cong\triangle ECQ$。
【问题探究】
(2) 在(1)的条件下，若$CG = 3$，$GQ = 5$，求$DQ$的长。
【拓展延伸】
(3) 如图②，当$CE = 2BE$时，点$P$在$BC$边上。若$\frac{CQ}{DQ}=\frac{1}{n}$，求$\frac{CG}{DG}$的值（用含$n$的代数式表示）。

答案： 21.
(1)由折叠的性质，得$\angle B = \angle AFE$，$BE = FE$。$\because$四边形$ABCD$是平行四边形，$\therefore AB// CD$。$\therefore\angle B = \angle PCG$。$\therefore\angle AFE = \angle PCG$。$\because\angle AFE = \angle QFG$，$\therefore\angle PCG = \angle QFG$。$\because\angle FGQ = \angle CGP$，$\therefore\angle CQE = \angle P$。$\because CE = BE$，$BE = EF$，$\therefore EF = EC$。又$\because\angle CFQ = \angle FEP$，$\therefore\triangle EFP\cong\triangle ECQ$。
(2)$\because\triangle EFP\cong\triangle ECQ$，$\therefore EQ = EP$。$\because EF = EC$，$\therefore FQ = CP$。$\because\angle FGQ = \angle CGP$，$\angle CQE = \angle P$，$\therefore\triangle FQG\cong\triangle CPG$。$\therefore FG = CG = 3$，$GQ = GP = 5$。由折叠的性质，得$AF = AB$。$\because$四边形$ABCD$是平行四边形，$\therefore AB// CD$，$AB = CD$。$\therefore\triangle CGP\sim\triangle BAP$。$\therefore\frac{CG}{BA}=\frac{PG}{PA}$。$\therefore\frac{3}{AB}=\frac{5}{AB + 3 + 5}$，解得$AB = 12$。$\therefore CD = 12$。$\therefore DQ = CD - CG - QG = 4$。
(3)延长$AD$，$EQ$交于点$M$。设$CQ = a$，$BE = b$。$\because\frac{CQ}{DQ}=\frac{1}{n}$，$CE = 2BE$，$\therefore DQ = an$，$EC = 2b$。$\therefore AB = CD=(n + 1)a$，$AD = 3b$。由折叠的性质，得$AF = AB=(n + 1)a$。$\because AD// BC$，即$DM// EC$，$\therefore\triangle DQM\sim\triangle CQE$。$\therefore\frac{DM}{CE}=\frac{DQ}{CQ}$，即$\frac{DM}{2b}=\frac{an}{a}=n$。$\therefore DM = 2bn$。$\because$四边形$ABCD$是平行四边形，$\therefore\angle B = \angle ADQ$。由折叠的性质，得$\angle AFE = \angle B$。$\because\angle AFQ + \angle AFE = 180^{\circ}$，$\therefore\angle AFQ + \angle ADQ = 180^{\circ}$。$\therefore\angle DAF + \angle DQF = 180^{\circ}$。$\because\angle EQC + \angle DQF = 180^{\circ}$，$\therefore\angle EQC = \angle DAF$。$\because AD// BC$，$\therefore\angle DAF = \angle FPE$。$\therefore\angle EQC = \angle FPE$。又$\because\angle FEP = \angle CEQ$，$\therefore\triangle FEP\sim\triangle CEQ$。$\therefore\frac{EF}{EC}=\frac{FP}{CQ}$，即$\frac{b}{2b}=\frac{FP}{a}$。$\therefore FP=\frac{1}{2}a$。$\because AD// BC$，$\therefore\triangle AMF\sim\triangle PEF$。$\therefore\frac{AM}{PE}=\frac{AF}{PF}$。$\therefore PE=\frac{AM· PF}{AF}=\frac{(3b + 2bn)·\frac{1}{2}a}{(n + 1)a}=\frac{3 + 2n}{2n + 2}b$。$\therefore CP = EC - EP = 2b-\frac{3 + 2n}{2n + 2}b=\frac{(2n + 1)b}{2n + 2}$。又$\because PC// AD$，$\therefore\triangle GPC\sim\triangle GAD$。$\therefore\frac{CG}{DG}=\frac{CP}{DA}=\frac{\frac{(2n + 1)b}{2n + 2}}{3b}=\frac{2n + 1}{6n + 6}$。