《2025年名校题库八年级数学上册北师大版》

2. (双流区期末)如图, 在$\triangle ABC$中,$\angle ACB = 90^{\circ}$,$AC = 8$,$BC = 6$,$D为AB$的中点, 点$E$,$F分别在直线BC$,$AC$上,$DF\bot DE$, 连接$EF$.
(1)如图 1, 当点$E与点B$重合时, 求$EF$的长;
(2)如图 2, 当点$F不与点A$重合时, 求证:$AF^{2}+BE^{2}= EF^{2}$;
(3)若$EC = 1$, 直接写出线段$CF$的长.

答案：

(1)解:$\because D$为$AB$的中点,$DF⊥DE,\therefore DF$垂直平分$AB,\therefore AF=EF$.
设$AF=EF=x$,在$Rt△CEF$中,由勾股定理,得$CF^{2}+EC^{2}=EF^{2},\therefore (8 - x)^{2}+6^{2}=x^{2}$,解得$x=\frac{25}{4},\therefore EF=\frac{25}{4}$.
(2)证明:如图1,过点A作$AG⊥AC$,交ED的延长线于点G,连接FG.

$\because D$为$AB$的中点,$\therefore AD=BD$.
$\because AG⊥AC,\therefore ∠GAC=∠ACB=90^{\circ },\therefore AG// BC,\therefore ∠AGD=∠BED$.
在$△AGD$和$△BED$中,$\begin{cases}∠AGD = ∠BED\\∠ADG = ∠BDE\\AD = BD\end{cases}$
$\therefore △AGD\cong △BED(AAS),\therefore BE=AG,DG=DE$.
$\because DF⊥DE,\therefore DF$是$GE$的垂直平分线,$\therefore GF=EF$.
$\because ∠GAF=90^{\circ },\therefore AF^{2}+AG^{2}=FG^{2},\therefore AF^{2}+BE^{2}=EF^{2}$.
(3)解:如图2,当点E在线段BC上时,作$BH// AC$,交FD的延长线于点H,连接EH.

与
(2)同理可得$△ADF\cong △BDH(AAS),\therefore BH=AF,DH=DF$.
$\because DE⊥DF,\therefore DE$是$HF$的垂直平分线,$\therefore EF=HE,\therefore CF^{2}+CE^{2}=AF^{2}+BE^{2}$.
$\because BC=6,EC=1,\therefore BE=5$.设$CF=m$,则$AF=8 - m,\therefore m^{2}+1^{2}=(8 - m)^{2}+5^{2}$,解得$m=\frac{11}{2},\therefore CF=\frac{11}{2}$.
如图3,当点E在BC的延长线上时,作$BG// AC$,交FD的延长线于点G,连接EG.

同理可得$CF^{2}+CE^{2}=AF^{2}+BE^{2}$.
易得$BE=7$.
设$CF=m$,则$AF=8 - m,\therefore m^{2}+1^{2}=(8 - m)^{2}+7^{2}$,解得$m=7,\therefore CF=7$.
综上所述,线段$CF$的长为7或$\frac{11}{2}$.

3. (成华区期末)如图, 在$\triangle ABC$中,$\angle ACB = 90^{\circ}$,$CA = CB$,$M是AB$的中点, 点$D在BM$上,$AE\bot CD$,$BF\bot CD$, 垂足分别为$E$,$F$, 连接$ME$,$MF$.
(1)求证:$CE = BF$;
(2)求证:$\triangle EFM$是等腰直角三角形;
(3)试判断线段$DE$,$DF$,$DM$之间有何数量关系, 写出你的结论并证明.

答案：

(1)证明:$\because ∠ACB=90^{\circ },\therefore ∠BCF+∠ACE=90^{\circ }$.
$\because AE⊥CD,BF⊥CD,\therefore ∠CEA=∠BFC=90^{\circ },\therefore ∠BCF+∠CBF=90^{\circ },\therefore ∠ACE=∠CBF$.
又$\because AC=CB,\therefore △CAE\cong △BCF(AAS),\therefore CE=BF$.
(2)证明:$\because △CAE\cong △BCF,\therefore AE=CF,BF=CE,\therefore AE - CE=CF - CE=EF$.
$\because M$是$AB$的中点,$CA=CB,∠ACB=90^{\circ },\therefore CM=\frac{1}{2}AB=BM=AM,CM⊥AB,\therefore ∠CMB=90^{\circ }$.
在$△BDF$和$△CDM$中,$∠BFD=∠CMD,∠BDF=∠CDM,\therefore ∠DBF=∠DCM$.
又$\because BF=CE,BM=CM,\therefore △BFM\cong △CEM(SAS),\therefore FM=EM,∠BMF=∠CME$,
$\therefore ∠BMF+∠DME=∠CME+∠DME=∠BMC=90^{\circ }$,即$∠EMF=90^{\circ },\therefore △EFM$为等腰直角三角形.
(3)解:$DE^{2}+DF^{2}=2DM^{2}$.证明如下:
设$AE$与$CM$交于点$N$,连接$DN$,如图所示.
由
(2)知$∠DBF=∠NCE$.
又$\because BF=CE,∠BFD=∠CEN=90^{\circ },\therefore △BFD\cong △CEN(ASA),\therefore DF=NE,BD=CN$.
$\because CM=BM,\therefore CM - CN=BM - BD$,即$DM=NM,\therefore △DMN$是等腰直角三角形,$\therefore DN^{2}=DM^{2}+NM^{2}=2DM^{2}$.
$\because AE⊥CD,\therefore ∠AED=90^{\circ }$.
在$Rt△DEN$中,由勾股定理,得$DN^{2}=DE^{2}+NE^{2},\therefore DN^{2}=DE^{2}+DF^{2},\therefore DE^{2}+DF^{2}=2DM^{2}$.