《2025年名校题库九年级数学全一册北师大版》

3. (嘉祥)如图,在平行四边形$ABCD$中,$AB= 6$,$AD= 9$,$\angle BAD的平分线交BC于点E$,交$DC的延长线于点F$,$BG\perp AE$,垂足为$G$,$BG= 4\sqrt{2}$,则$\triangle CEF$的周长为______8.

答案： 8【解析】在$□ ABCD$中，$\because AB = CD = 6$，$AD = BC = 9$，$\angle BAD$的平分线交$BC$于点$E$，$\therefore \angle BAF = \angle DAF$。$\because AB// DF$，$\therefore \angle BAF = \angle F$，$\therefore \angle F = \angle DAF$，$\therefore \triangle ADF$是等腰三角形，$AD = DF = 9$，$\therefore FC = 9 - 6 = 3$。$\because AD// BC$，$\therefore \angle CEF = \angle DAF = \angle F$，$\therefore \triangle EFC$是等腰三角形，且$FC = CE = 3$，$\therefore AB = BE = 6$。又$\because BG\perp AE$，$\therefore AG = GE$。在$Rt\triangle ABG$中，由勾股定理，得$AG = \sqrt{AB^{2} - BG^{2}} = 2$，$\therefore AE = 2AG = 4$，$\therefore \triangle ABE$的周长等于16。$\because AB// CF$，$\therefore \triangle CEF\backsim\triangle BEA$，且相似比为$1:2$，$\therefore \triangle CEF$的周长为8。

4. (温江区二诊)如图,在$\triangle ABC$中,$D$,$E分别是边AB$,$AC$的中点,$DF过EC的中点G并与BC的延长线交于点F$,$BE与DF交于点O$.若$\triangle ADE的面积为S$,则四边形$BOGC$的面积为______$\frac{7}{4}S$.

答案： $\frac{7}{4}S$

5. (青羊区一诊)如图,在$Rt\triangle ABC$中,$\angle BAC= 90^{\circ}$,$AB= 4$,将$\triangle ABC绕点A逆时针旋转得到\triangle AB_{1}C_{1}$,$B_{1}C_{1}交BC于点D$,$AB_{1}交BC于点E$,连接$AD$,当$AE平分\angle BAD$时,$AE= 3$,则$BD= $______3.5.

答案： 3.5【解析】$\because \angle B_{1} = \angle B$，$\angle BEA = \angle B_{1}ED$，$\therefore \angle B_{1}DE = \angle BAE$，$\therefore \angle B_{1}DE = \angle DAE$。又$\because \angle B_{1} = \angle B_{1}$，$\therefore \triangle B_{1}DE\backsim\triangle B_{1}AD$，$\therefore \frac{B_{1}E}{DB_{1}} = \frac{DB_{1}}{AB_{1}}$，$\therefore \frac{1}{DB_{1}} = \frac{DB_{1}}{4}$，$\therefore DB_{1} = 2$。$\because \angle B_{1} = \angle B$，$\angle BEA = \angle B_{1}ED$，$\therefore \triangle B_{1}DE\backsim\triangle BAE$，$\therefore \frac{DE}{AE} = \frac{EB_{1}}{EB} = \frac{DB_{1}}{AB} = \frac{1}{2}$，$\therefore DE = \frac{1}{2}AE = \frac{3}{2}$，$EB = 2EB_{1} = 2$，$\therefore DB = DE + BE = 3.5$。

6. (成华区一诊)如图,在正方形$ABCD$中,延长$AD到点E$,连接$CE$,过点$A作AH\perp CE$,垂足为$H$,$AH交CD于点F$,交$BD于点G$.
(1)求证：$AF= CE$；
(2)若$\frac{DE}{AD}= \frac{2}{5}$,求$\frac{GF}{FH}$的值.

(1)证明：$\because AH\perp CE$，$\therefore \angle AHE = 90^{\circ}$。
$\because$四边形$ABCD$是正方形，点$E$在$AD$的延长线上，$\therefore AD = CD$，$\angle ADF = \angle CDE = 90^{\circ}$，
$\therefore \angle DAF = \angle DCE = 90^{\circ} - \angle E$。
在$\triangle ADF$和$\triangle CDE$中，$\begin{cases}\angle ADF = \angle CDE,\\AD = CD,\\\angle DAF = \angle DCE,\end{cases}$ $\therefore \triangle ADF\cong\triangle CDE(ASA)$，$\therefore AF = CE$。
(2)解：由(1)得$\triangle ADF\cong\triangle CDE$，$\therefore DF = DE$。
设$DF = DE = 2m$。$\because \frac{DE}{AD} = \frac{2}{5}$，$\therefore BA = DC = AD = \frac{5}{2}DE = \frac{5}{2}× 2m = 5m$，
$\therefore CF = DC - DF = 5m - 2m = 3m$，$AF = \sqrt{AD^{2} + DF^{2}} = \sqrt{(5m)^{2} + (2m)^{2}} = \sqrt{29}m$。
$\because DF// BA$，$\therefore \triangle DFG\backsim\triangle BAG$，$\therefore \frac{GF}{GA} = \frac{DF}{BA} = \frac{2m}{5m} = \frac{2}{5}$，
$\therefore GF = \frac{2}{2 + 5}AF = \frac{2}{7}AF = \frac{2}{7}×\sqrt{29}m = \frac{2\sqrt{29}}{7}m$。
$\because \angle CHF = \angle ADF = 90^{\circ}$，$\angle CFH = \angle AFD$，
$\therefore \triangle CHF\backsim\triangle ADF$，$\therefore \frac{FH}{DF} = \frac{CF}{AF}$，
$\therefore FH = \frac{CF\cdot DF}{AF} = \frac{3m\cdot 2m}{\sqrt{29}m} = \frac{6\sqrt{29}}{29}m$，
$\therefore \frac{GF}{FH} = \frac{\frac{2\sqrt{29}}{7}m}{\frac{6\sqrt{29}}{29}m} = \frac{29}{21}$。

答案：
(1)证明：$\because AH\perp CE$，$\therefore \angle AHE = 90^{\circ}$。
$\because$四边形$ABCD$是正方形，点$E$在$AD$的延长线上，$\therefore AD = CD$，$\angle ADF = \angle CDE = 90^{\circ}$，
$\therefore \angle DAF = \angle DCE = 90^{\circ} - \angle E$。
在$\triangle ADF$和$\triangle CDE$中，$\begin{cases}\angle ADF = \angle CDE,\\AD = CD,\\\angle DAF = \angle DCE,\end{cases}$ $\therefore \triangle ADF\cong\triangle CDE(ASA)$，$\therefore AF = CE$。
(2)解：由
(1)得$\triangle ADF\cong\triangle CDE$，$\therefore DF = DE$。
设$DF = DE = 2m$。$\because \frac{DE}{AD} = \frac{2}{5}$，$\therefore BA = DC = AD = \frac{5}{2}DE = \frac{5}{2}\times 2m = 5m$，
$\therefore CF = DC - DF = 5m - 2m = 3m$，$AF = \sqrt{AD^{2} + DF^{2}} = \sqrt{(5m)^{2} + (2m)^{2}} = \sqrt{29}m$。
$\because DF// BA$，$\therefore \triangle DFG\backsim\triangle BAG$，$\therefore \frac{GF}{GA} = \frac{DF}{BA} = \frac{2m}{5m} = \frac{2}{5}$，
$\therefore GF = \frac{2}{2 + 5}AF = \frac{2}{7}AF = \frac{2}{7}\times\sqrt{29}m = \frac{2\sqrt{29}}{7}m$。
$\because \angle CHF = \angle ADF = 90^{\circ}$，$\angle CFH = \angle AFD$，
$\therefore \triangle CHF\backsim\triangle ADF$，$\therefore \frac{FH}{DF} = \frac{CF}{AF}$，
$\therefore FH = \frac{CF\cdot DF}{AF} = \frac{3m\cdot 2m}{\sqrt{29}m} = \frac{6\sqrt{29}}{29}m$，
$\therefore \frac{GF}{FH} = \frac{\frac{2\sqrt{29}}{7}m}{\frac{6\sqrt{29}}{29}m} = \frac{29}{21}$。

7. (高新区一诊)如图,在四边形$ABCD$中,对角线$AC$,$BD交于点O$,$AD// BC$,$AC垂直平分BD$.
(1)求证：四边形$ABCD$是菱形；
(2)已知四边形$ABCD的面积为20$,$AB= 5$,点$E在线段OD$上,$ED= 2OE$,射线$AE交CD于点F$,交$BC的延长线于点G$,求线段$EF$的长.

答案：

(1)证明：$\because AC$垂直平分$BD$，$\therefore BO = DO$，$AB = AD$。
$\because AD// BC$，$\therefore \angle ADO = \angle CBO$。
在$\triangle ADO$与$\triangle CBO$中，$\begin{cases}\angle ADO = \angle CBO,\\OD = OB,\\\angle AOD = \angle COB,\end{cases}$
$\therefore \triangle ADO\cong\triangle CBO(ASA)$，$\therefore AD = CB$。
$\because AD// BC$，$\therefore$四边形$ABCD$是平行四边形。
$\because AB = AD$，$\therefore$四边形$ABCD$是菱形。
(2)解：如图，过点$A$作$AH\perp BC$于点$H$。
$\because$四边形$ABCD$是菱形，$\therefore AB = BC = 5$。
$\because$四边形$ABCD$的面积为20，$\therefore BC\cdot AH = 5AH = 20$，$\therefore AH = 4$，
$\therefore BH = \sqrt{AB^{2} - AH^{2}} = 3$，$\therefore CH = BC - BH = 2$，$\therefore AC = \sqrt{AH^{2} + CH^{2}} = 2\sqrt{5}$，
$\therefore AO = CO = \frac{1}{2}AC = \sqrt{5}$，$\therefore BO = OD = \sqrt{AB^{2} - AO^{2}} = 2\sqrt{5}$。
$\because ED = 2OE$，$\therefore OE = \frac{2\sqrt{5}}{3}$，$OB = OD = 3OE$，
$\therefore AE = \sqrt{AO^{2} + OE^{2}} = \frac{\sqrt{65}}{3}$。
$\because AB// DF$，$\therefore \triangle DEF\backsim\triangle BEA$，$\therefore \frac{AE}{EF} = \frac{BE}{DE}$，
$\therefore \frac{\frac{\sqrt{65}}{3}}{EF} = \frac{3OE + OE}{2OE}$，$\therefore EF = \frac{\sqrt{65}}{6}$。