《2026年通城学典全国中考试题分类精粹中考数学》

14. (2025·成都)如图，在$\triangle ABC$中，$AB = AC$，点$D$在$AC$边上，$AD = 3$，$CD = 2$，$\angle CBD = 45^{\circ}$，则$\tan\angle ACB$的值为4；点$E$在$BC$的延长线上，连接$DE$，若$\angle E=\angle ABD$，则$CE$的长为$\frac{2\sqrt{17}}{3}$。

答案：
14.$4\frac{2\sqrt{17}}{3}$ 解析：如图，过点$A$作$AH\perp BC$于点$H$，过点$D$作$DG\perp BC$于点$G$，过点$D$作$DF\perp AH$于点$F$，则四边形$DFHG$为矩形，$\therefore DG = FH$，$DF = HG$，$DF// HG$，$DG// AH$。$\because\angle DBC = 45^{\circ}$，$\therefore\triangle BDG$为等腰直角三角形。
$\therefore BG = DG$。$\because AB = AC$，$\therefore BH = CH$，$\angle ABC = \angle ACB$。
$\because DF// BC$，$\therefore\triangle ADF\sim\triangle ACH$。$\therefore\frac{DF}{CH}=\frac{AD}{AC}=\frac{AD}{AD + CD}=\frac{3}{5}$。$\therefore$设$DF = 3x$，$CH = 5x$，则$HG = DF = 3x$，$BH = CH = 5x$。$\therefore DG = BG = BH + HG = 8x$，$CG = CH - HG = 2x$。$\therefore BD = 8\sqrt{2}x$。$\therefore$在$Rt\triangle CGD$中，$\tan\angle ACB=\frac{DG}{CG}=\frac{8x}{2x}=4$。在$Rt\triangle CGD$中，由勾股定理，得$CG^{2}+DG^{2}=CD^{2}$，即$(2x)^{2}+(8x)^{2}=2^{2}$，解得$x = \frac{\sqrt{17}}{17}$（负值舍去）。$\therefore BD = 8\sqrt{2}x=\frac{8\sqrt{34}}{17}$，$BC = 2CH = 10x=\frac{10\sqrt{17}}{17}$。
$\because\angle E = \angle ABD$，$\angle ACB = \angle E + \angle CDE$，$\angle ABC = \angle ABD + \angle CBD$，$\angle ABC = \angle ACB$，$\therefore\angle CDE = \angle CBD = 45^{\circ}$。又$\because\angle E = \angle E$，$\therefore\triangle DEC\sim\triangle BED$。
$\therefore\frac{DE}{BE}=\frac{CE}{DE}=\frac{CD}{DB}=\frac{2}{8\sqrt{34}}=\frac{\sqrt{34}}{8}$。$\therefore DE = \frac{8}{\sqrt{34}}CE$，
$DE^{2}=BE· CE=(BC + CE)· CE$。$\therefore(\frac{8}{\sqrt{34}}CE)^{2}=(\frac{10\sqrt{17}}{17}+CE)· CE$，解得$CE = 0$（舍去）或$CE=\frac{2\sqrt{17}}{3}$。

15. (2025·广安)已知$\triangle ABC$的面积是$1$。
(1) 如图①，若$D$，$E$分别是边$BC$和$AC$的中点，$AD$与$BE$相交于点$F$，则四边形$CDFE$的面积为$\frac{1}{3}$；
(2) 如图②，若$M$，$N$分别是边$BC$和$AC$上距离点$C$最近的六等分点，$AM$与$BN$相交于点$G$，则四边形$CMGN$的面积为$\frac{1}{21}$。

答案： 15.
(1)$\frac{1}{3}$
(2)$\frac{1}{21}$ 解析：
(1)连接$DE$。$\because D$，$E$分别是边$BC$和$AC$的中点，$\therefore DE$是$\triangle ABC$的中位线。$\therefore DE// AB$，$DE = \frac{1}{2}AB$。$\therefore\triangle CDE\sim\triangle CBA$。$\therefore\frac{S_{\triangle CDE}}{S_{\triangle CBA}}=(\frac{DE}{AB})^{2}=\frac{1}{4}$。$\because\triangle ABC$的面积是$1$，$\therefore S_{\triangle CDE}=\frac{1}{4}$。$\because D$是$BC$的中点，$\therefore S_{\triangle BDE}=S_{\triangle CDE}=\frac{1}{4}$。$\because DE// AB$，
$\therefore\triangle DEF\sim\triangle ABF$。$\therefore\frac{EF}{BF}=\frac{DE}{AB}=\frac{1}{2}$。$\therefore BF = 2EF$。$\therefore BE = BF + EF = 3EF$。$\therefore\frac{S_{\triangle DEF}}{S_{\triangle BDE}}=\frac{EF}{BE}=\frac{EF}{3EF}=\frac{1}{3}$。$\therefore S_{\triangle DEF}=\frac{1}{12}$。$\therefore S_{四边形CDFE}=S_{\triangle CDE}+S_{\triangle DEF}=\frac{1}{4}+\frac{1}{12}=\frac{1}{3}$。
(2)连接$MN$。$\because M$，$N$分别是边$BC$和$AC$上距离点$C$最近的六等分点，$\therefore CM = \frac{1}{6}BC$，$CN = \frac{1}{6}AC$。
$\because\angle C = \angle C$，$\therefore\triangle CMN\sim\triangle CBA$。$\therefore\frac{S_{\triangle CMN}}{S_{\triangle ABC}}=(\frac{CM}{BC})^{2}=\frac{1}{36}$。$\because\frac{MN}{AB}=\frac{CM}{BC}=\frac{1}{6}$，$\angle CMN = \angle CBA$。$\therefore MN// AB$。$\because\triangle ABC$的面积是$1$，
$\therefore S_{\triangle CMN}=\frac{1}{36}$。$\because M$是边$BC$上距离点$C$最近的六等分点，
$\therefore\frac{BM}{CM}=5$。$\therefore\frac{S_{\triangle BMN}}{S_{\triangle CMN}}=\frac{BM}{CM}=5$。$\therefore S_{\triangle BMN}=\frac{5}{36}$。$\because MN// AB$，$\therefore\triangle MNG\sim\triangle ABG$。$\therefore\frac{NG}{BG}=\frac{MN}{AB}=\frac{1}{6}$。$\therefore BG = 6NG$。$\therefore BN = BG + NG = 7NG$。$\therefore\frac{S_{\triangle MNG}}{S_{\triangle BMN}}=\frac{NG}{BN}=\frac{1}{7}$。$\therefore S_{\triangle MNG}=\frac{1}{7}S_{\triangle BMN}=\frac{5}{252}$。$\therefore S_{四边形CMGN}=S_{\triangle MNG}+S_{\triangle CMN}=\frac{5}{252}+\frac{1}{36}=\frac{1}{21}$。

16. (2025·陕西)如图，点$O$在$\triangle ABC$的边$AC$上，以$OC$为半径的$\odot O$与$AB$相切于点$D$，与$BC$相交于点$E$，$EF$为$\odot O$的直径，$FD$与$AC$相交于点$G$，$\angle F = 45^{\circ}$。
(1) 求证：$AB = AC$；
(2) 若$\sin A=\frac{3}{5}$，$AB = 8$，求$DG$的长。

答案： 16.
(1)连接$OD$。$\because\angle F = 45^{\circ}$，$\therefore\angle DOE = 2\angle F = 90^{\circ}$。$\because\odot O$与$AB$相切于点$D$，$\therefore OD\perp AB$。$\therefore\angle ODA = \angle DOE = 90^{\circ}$。$\therefore AB// OE$。$\therefore\angle B = \angle OEC$。$\because OC = OE$，
$\therefore\angle C = \angle OEC$。$\therefore\angle B = \angle C$。$\therefore AB = AC$。
(2)由
(1)，
得$\angle ODA = \angle DOF = 90^{\circ}$。$\because$在$Rt\triangle ODA$中，$\sin A=\frac{OD}{OA}=\frac{3}{5}$，$\therefore OA=\frac{5}{3}OD$。$\because OF = OC = OD$，$OA + OC = AC = AB = 8$，$\angle DOF = 90^{\circ}$，$\therefore\frac{5}{3}OD + OD = 8$。$\therefore OF = OD = 3$。
$\therefore OA=\frac{5}{3}×3 = 5$，$DF = \sqrt{OF^{2}+OD^{2}}=\sqrt{2}OF = 3\sqrt{2}$。$\therefore AD = \sqrt{OA^{2}-OD^{2}}=\sqrt{5^{2}-3^{2}} = 4$。$\because AD// OF$，
$\therefore\triangle AGD\sim\triangle OGF$。$\therefore\frac{DG}{FG}=\frac{AD}{OF}=\frac{4}{3}$。$\therefore DG=\frac{4}{4 + 3}DF=\frac{4}{7}DF=\frac{4}{7}×3\sqrt{2}=\frac{12\sqrt{2}}{7}$。$\therefore DG$的长是$\frac{12\sqrt{2}}{7}$。

17. (2025·南通)如图，在矩形$ABCD$中，对角线$AC$，$BD$相交于点$O$，$M$是$BC$的中点，$DM$交$AC$于点$G$。
(1) 求证：$AG = 2CG$。
(2) 设$\angle BCD$，$\angle BDC$的平分线相交于点$I$。
① 当$AB = 6$，$BC = 8$时，求点$I$到$BC$的距离；
② 若$AB + AC = 2BC$，作直线$GI$分别交$BD$，$CD$于$E$，$F$两点，求$\frac{EF}{BC}$的值。

答案：
17.
(1)$\because$四边形$ABCD$为矩形，$\therefore AD = BC$，$AD// BC$。
$\therefore\triangle ADG\sim\triangle CMG$。$\therefore\frac{AG}{CG}=\frac{AD}{CM}$。$\because M$是$BC$的中点，
$\therefore CM=\frac{1}{2}BC=\frac{1}{2}AD$。$\therefore\frac{AD}{CM}=2$。$\therefore\frac{AG}{CG}=2$。$\therefore AG = 2CG$。
(2)①如图①，过点$I$作$IH\perp CD$于点$H$，作$IR\perp BC$于点$R$，作$IQ\perp BD$于点$Q$，连接$IB$。$\because\angle BCD$，$\angle BDC$的平分线相交于点$I$，$\therefore IH = IR = IQ$。设$IH = IR = IQ = x$。$\because$四边形$ABCD$为矩形，$\therefore\angle BCD = 90^{\circ}$，$CD = AB = 6$。
又$\because BC = 8$，$\therefore$在$Rt\triangle BCD$中，$BD=\sqrt{BC^{2}+CD^{2}}=10$。
$\because S_{\triangle BCD}=S_{\triangle IBC}+S_{\triangle ICD}+S_{\triangle IBD}$，$\therefore\frac{1}{2}×6×8=\frac{1}{2}×8× IR+\frac{1}{2}×6× IH+\frac{1}{2}×10× IQ$，即$\frac{1}{2}×6×8=\frac{1}{2}×8x+\frac{1}{2}×6x+\frac{1}{2}×10x$，解得$x = 2$。$\therefore$点$I$到$BC$的距离为$2$。
②如图②，过点$G$作$GK\perp BC$于点$K$，过点$I$作$IN\perp BC$于点$N$。设$AB = m$，$AC = n$。$\because AB + AC = 2BC$，$\therefore BC=\frac{m + n}{2}$。
在$Rt\triangle ABC$中，由勾股定理，得$AB^{2}+BC^{2}=AC^{2}$，
即$m^{2}+(\frac{m + n}{2})^{2}=n^{2}$，解得$m_{1}=-n$（舍去），$m_{2}=\frac{3}{5}n$。由
(1)知，$AG = 2CG$，$\therefore GC=\frac{1}{3}AC=\frac{1}{3}n$。易知$\triangle CGK\sim\triangle CAB$，$\frac{GK}{AB}=\frac{CG}{CA}$，$\therefore GK=\frac{1}{5}n$。$\therefore BC=\frac{4}{5}n$。同①，可得$IN=\frac{1}{5}n = GK$，$\therefore$易得四边形$GKNI$为矩形。$\therefore EF// BC$。又$\because AD// BC$，$\therefore\frac{EF}{BC}=\frac{DG}{DM}=\frac{AG}{AC}=\frac{2}{3}$。