8. (2025·宁波期中) 如图,$AB// DC$,M 和 N 分别是 AD 和 BC 的中点,连接 CM,DN 并延长,分别交 AB 于 Q,P,若四边形 ABCD 的面积为 $24\ \mathrm{cm^2}$,则 $S_{△ QPO}-S_{△ CDO}=\_\_\_\_\_\_\mathrm{cm^2}$.

8. 24 解析:$\because AB // DC$,$\therefore ∠ Q = ∠ DCM$,$∠ P = ∠ CDN$.$\because M$和$N$分别是$AD$和$BC$的中点,$\therefore AM = DM$,$BN = CN$.$\because ∠ AMQ = ∠ DMC$,$\therefore △ AQM ≌ △ DCM(\mathrm{AAS})$,同理可得$△ BPN ≌ △ CDN$,$\therefore S_{△ AQM} = S_{△ DCM}$,$S_{△ BPN} = S_{△ CDN}$,$\therefore S_{△ QPO} = S_{△ AQM} + S_{△ BPN} + S_{\mathrm{五边形}ABNOM} = S_{△ CDM} + S_{△ CDN} + S_{\mathrm{五边形}ABNOM} = S_{\mathrm{梯形}ABCD} + S_{△ CDO}$,$\therefore S_{△ QPO} - S_{△ CDO} = S_{\mathrm{梯形}ABCD} = 24\ \mathrm{cm}^2$.

9. (2026·北京期中) 如图,在$△ ABC$中,$D$为边$BC$上一点,$E$为边$BA$上一点,且$AE=CD$,连接$AD$,$F$为$AD$的中点.连接$EF$并延长,交$AC$于点$G$,在$FG$上截取点$H$,使$FH=FE$,连接$GD$,若$HG=CG$.
求证:(1)$△ AEF ≌ △ DHF$;
(2)$∠ B=2∠ GDC$.

9. (1)$\because F$是$AD$的中点,$\therefore AF = DF$.在$△ AEF$和$△ DHF$中,$\begin{cases} AF = DF,\\ ∠ AFE = ∠ DFH,\\ FE = FH, \end{cases}$$\therefore △ AEF ≌ △ DHF(\mathrm{SAS})$.
(2)$\because △ AEF ≌ △ DHF$,$\therefore AE = DH$,$∠ AEF = ∠ DHF$,$\therefore AB // DH$,$\therefore ∠ B = ∠ HDC$.$\because AE = CD$,$\therefore DH = CD$.在$△ HGD$和$△ CGD$中,$\begin{cases} DH = DC,\\ HG = CG,\\ DG = DG, \end{cases}$$\therefore △ HGD ≌ △ CGD(\mathrm{SSS})$,$\therefore ∠ HDG = ∠ CDG$,$\therefore ∠ HDC = 2∠ GDC$,$\therefore ∠ B = 2∠ GDC$.

10. (2025·达州期末) 如图, 在 $△ ABC$ 中, $AB = AC,P,Q$ 分别为边 $AB,AC$ 上两个动点, 在运动过程中始终保持 $AP+AQ=AB$, 连接 $BQ$ 和 $CP$, 当 $BQ+CP$ 值达到最小时, $\dfrac{AP}{AQ}$ 的值为.

10. 1 解析:如图,过点$B$作$BE // AC$,且$BE = AC$,在$BA$上截取$BH = AP$,连接$CH$,$HE$,$\because AB = AC$,$AP + AQ = AB$,$AB = AP + BP$,$AC = AQ + CQ$,$\therefore AQ = BP$,$CQ = AP = BH$.$\because AC // BE$,$\therefore ∠ A = ∠ EBH$.在$△ ACP$和$△ BEH$中,$\begin{cases} AC = BE,\\ ∠ A = ∠ EBH,\\ AP = BH, \end{cases}$$\therefore △ ACP ≌ △ BEH(\mathrm{SAS})$,$\therefore CP = HE$.$\because AB = AC$,$\therefore ∠ ACB = ∠ ABC$.在$△ CBQ$和$△ BCH$中,$\begin{cases} CB = BC,\\ ∠ BCQ = ∠ CBH,\\ CQ = BH, \end{cases}$$\therefore △ CBQ ≌ △ BCH(\mathrm{SAS})$,$\therefore CH = BQ$,$\therefore BQ + CP = CH + HE$,$\therefore$当$C$,$E$,$H$三点共线时,$BQ + CP$有最小值,此时,$\because AC // BE$,$\therefore ∠ A = ∠ EBA$,$∠ ACH = ∠ BEH$.又$\because AC = BE$,$\therefore △ ACH ≌ △ BEH(\mathrm{ASA})$,$\therefore AH = BH$,$\therefore$点$H$是$AB$的中点,$\therefore AP = BH = \dfrac{1}{2}AB$,$\therefore$点$P$与点$H$重合,$\therefore BP = BH = AQ = AP$,$\therefore \dfrac{AP}{AQ} = 1$.

11. 阅读与证明:在一个三角形中,如果有两个角相等,那么这两个角所对的边也相等. 如图①, 在 $△ ABC$ 中, 如果 $∠ B = ∠ C$, 那么 $AB = AC$,这一结论可以说明如下:
解: 过点 $A$ 作 $AD ⊥ BC$ 于 $D$, 则 $∠ ADB = ∠ ADC=90°$, 在 $△ ABD$ 和 $△ ACD$ 中,
$\begin{cases}∠ B = ∠ C, \\∠ ADB = ∠ ADC, \therefore △ ABD ≌ △ ACD, \\AD = AD,\end{cases}$
$\therefore AB=AC.$
(1)请你仿照上述方法在图②中再选一种方法说明以上结论.
(2)操作:如图③,点 $O$ 为线段 $MN$ 的中点,直线 $PQ$ 与 $MN$ 相交于点 $O$,过点 $M,N$ 作一组平行线分别与 $PQ$ 交于点 $M',N'$,则线段$MM'$一定等于 $NN'$. 试证明此结论.
(3)探究:如图④,在四边形 $ABCD$ 中, $AB // DC$,$E$ 为 $BC$ 边的中点, $∠ BAE = ∠ EAF$,$AF$与 $DC$ 的延长线相交于点 $F$. 试探究线段 $AB$与 $AF,CF$ 之间的等量关系,并说明你的结论.


视频讲题

11. (1)如图①,作$∠ BAC$的平分线$AD$,交$BC$于$D$,则$∠ BAD = ∠ CAD$,在$△ ABD$和$△ ACD$中,$\begin{cases} ∠ BAD = ∠ CAD,\\ ∠ B = ∠ C,\\ AD = AD, \end{cases}$$\therefore △ ABD ≌ △ ACD(\mathrm{AAS})$,$\therefore AB = AC$.


(2)如图②,$\because MM' // NN'$,$\therefore ∠ MM'N' = ∠ NN'M'$.$\because$点$O$为线段$MN$的中点,$\therefore OM = ON$.在$△ MOM'$和$△ NON'$中,$\begin{cases} ∠ MM'N' = ∠ NN'M',\\ ∠ MOM' = ∠ NON',\\ OM = ON, \end{cases}$$\therefore MM' = NN'$.
(3)如图③,连接$FE$并延长交$AB$于$G$,$\because AB // DC$,$\therefore ∠ B = ∠ ECF$.$\because E$为$BC$边的中点,$\therefore BE = CE$.在$△ BEG$和$△ CEF$中,$\begin{cases} ∠ B = ∠ ECF,\\ BE = CE,\\ ∠ BEG = ∠ CEF, \end{cases}$$\therefore EF = EG$,$BG = CF$.延长$AE$到$H$,使$AE = EH$,连接$FH$.在$△ AEG$和$△ HEF$中,$\begin{cases} AE = HE,\\ ∠ AEG = ∠ HEF,\\ EG = EF, \end{cases}$$\therefore △ AEG ≌ △ HEF\ (\mathrm{SAS})$,$\therefore AG = HF$,$∠ BAE = ∠ H$.$\because ∠ BAE = ∠ EAF$,$\therefore ∠ H = ∠ EAF$.由"阅读与证明"中的结论可得$AF = HF$,$\therefore AG = AF$.$\because AB = AG + BG$,$\therefore AB = AF + CF$.