8.(2024春·无锡期中改编)如图，$CB⊥AD$，$AE⊥CD$，垂足分别为$B$，$E$，且$AE$，$BC$相交于点$F$。若$AB=BC=16$，$CF=8$，连接$DF$，求图中阴影部分的面积。

8.解:$\because CB⊥ AD,AE⊥ CD$,
$\therefore ∠ABF=∠CBD=90°,∠FEC=90°$.
$\because ∠AFB=∠EFC,\therefore ∠A=∠C$.
在$△ ABF$和$△ CBD$中,$\begin{cases} ∠ABF=∠CBD,\\ AB=CB,\\ ∠A=∠C, \end{cases}$
$\therefore △ ABF≌△ CBD(\mathrm{ASA}),\therefore BF=BD$.
$\because BF=BC-CF=16-8=8,\therefore BD=8$,
$\therefore$阴影部分的面积$=\frac{1}{2}· FC· BD=\frac{1}{2}×8×8=32$.

9.如图,在$Rt△ ABC$中,$∠ ACB=90°$,过点$C$作$CD⊥ AB$,垂足为$D$.在射线$CD$上截取$CE=CA$,过点$E$作$EF⊥ CE$,交$CB$的延长线于点$F$.
(1)求证:$△ ABC≌△ CFE$;
(2)若$AB=9$,$EF=5$,求$BF$的长.

9.(1)证明:$\because EF⊥ CE,\therefore ∠E=90°$.
$\because ∠ACB=∠ADC=90°,\therefore ∠A=∠ECF=90°-∠ACE$.
在$△ ABC$和$△ CFE$中,$\begin{cases} ∠A=∠ECF,\\ CA=EC,\\ ∠ACB=∠E=90°, \end{cases}$
$\therefore △ ABC≌△ CFE(\mathrm{ASA})$.
(2)解:$\because △ ABC≌△ CFE,\therefore CF=AB=9,CB=EF=5$,
$\therefore BF=CF-CB=9-5=4$.

10.如图,在四边形ABCD中,AD//BC,E为CD的中点,连接AE,BE,BE⊥AE,延长AE交BC的延长线于点F.AD=2 cm,BC=5 cm.
(1)求证:AD=FC;
(2)求AB的长.

10.(1)证明:$\because AD// BC,\therefore ∠ADE=∠FCE$.
$\because E$是$CD$的中点,$\therefore DE=CE$.
在$△ ADE$和$△ FCE$中,$\begin{cases} ∠ADE=∠FCE,\\ DE=CE,\\ ∠AED=∠FEC, \end{cases}$
$\therefore △ ADE≌△ FCE(\mathrm{ASA}),\therefore AD=FC$.
(2)解:$\because △ ADE≌△ FCE,\therefore AE=EF$.
$\because BE⊥ AE,\therefore ∠AEB=∠FEB=90°$.
在$△ ABE$和$△ FBE$中,$\begin{cases} AE=FE,\\ ∠AEB=∠FEB,\\ BE=BE, \end{cases}$
$\therefore △ ABE≌△ FBE(\mathrm{SAS}),\therefore AB=BF$.
$\because AD=2\ \mathrm{cm},BC=5\ \mathrm{cm}$,
$\therefore AB=BF=BC+CF=BC+AD=5+2=7(\mathrm{cm})$.

11.如图,在四边形ABCD中,AD//BC,点E,F分别在AD,BC上,过点A,C分别作EF的垂线,垂足分别为G,H,且EH=FG.
(1)求证:△AGE≌△CHF;
(2)连接AC,线段GH与AC是否互相平分?请说明理由.

11.(1)证明:$\because EH=FG,\therefore EH-EF=FG-EF$,
即$FH=EG$.
$\because AG⊥ EF,CH⊥ EF,\therefore ∠G=∠H=90°$.
$\because AD// BC,\therefore ∠DEF=∠BFE$.
$\because ∠AEG=∠DEF,∠CFH=∠BFE$,
$\therefore ∠AEG=∠CFH$.
在$△ AGE$和$△ CHF$中,$\begin{cases} ∠G=∠H,\\ EG=FH,\\ ∠AEG=∠CFH, \end{cases}$
$\therefore △ AGE≌△ CHF(\mathrm{ASA})$.
(2)解:线段$GH$与$AC$互相平分.理由如下:
如答图,连接$AC$交$GH$于点$O$.
$\because △ AGE≌△ CHF,\therefore AG=CH$.
$\because ∠G=∠H=90°,\therefore AG// CH,\therefore ∠OAG=∠OCH$.
在$△ OAG$和$△ OCH$中,$\begin{cases} ∠G=∠H,\\ AG=CH,\\ ∠OAG=∠OCH, \end{cases}$
$\therefore △ OAG≌△ OCH(\mathrm{ASA}),\therefore OA=OC,OG=OH$,
$\therefore$线段$GH$与$AC$互相平分.