5. 如图,在四边形ABCD中,∠B+∠ADC=180°,AB=AD,E,F分别是边BC,CD延长线上的点,且∠EAF=$\frac{1}{2}$∠BAD,求证:EF=BE−FD.
答案：
5.如图,在$BE$上截取$BG$,使$BG=DF$,连接$AG$.
$\because ∠ B + ∠ ADC = 180°,$
$∠ ADF + ∠ ADC = 180°,$
$\therefore ∠ B = ∠ ADF.$
在$△ ABG$和$△ ADF$中,
$\begin{cases} AB=AD, \\ ∠ B=∠ ADF, \\ BG=DF, \end{cases}$
$\therefore △ ABG ≌ △ ADF (\mathrm{SAS}),$
$\therefore ∠ BAG = ∠ DAF, AG=AF,$
$\therefore ∠ BAG + ∠ EAD = ∠ DAF + ∠ EAD = ∠ EAF = \frac{1}{2}∠ BAD, \therefore ∠ GAE = ∠ EAF.$
在$△ AEG$和$△ AEF$中,$\begin{cases} AG=AF, \\ ∠ GAE=∠ FAE, \\ AE=AE, \end{cases}$
$\therefore △ AEG ≌ △ AEF (\mathrm{SAS}),$
$\therefore EG=EF. \because EG=BE-BG, \therefore EF=BE-FD.$
5.如图,在$BE$上截取$BG$,使$BG=DF$,连接$AG$.
$\because ∠ B + ∠ ADC = 180°,$
$∠ ADF + ∠ ADC = 180°,$
$\therefore ∠ B = ∠ ADF.$
在$△ ABG$和$△ ADF$中,
$\begin{cases} AB=AD, \\ ∠ B=∠ ADF, \\ BG=DF, \end{cases}$
$\therefore △ ABG ≌ △ ADF (\mathrm{SAS}),$
$\therefore ∠ BAG = ∠ DAF, AG=AF,$
$\therefore ∠ BAG + ∠ EAD = ∠ DAF + ∠ EAD = ∠ EAF = \frac{1}{2}∠ BAD, \therefore ∠ GAE = ∠ EAF.$
在$△ AEG$和$△ AEF$中,$\begin{cases} AG=AF, \\ ∠ GAE=∠ FAE, \\ AE=AE, \end{cases}$
$\therefore △ AEG ≌ △ AEF (\mathrm{SAS}),$
$\therefore EG=EF. \because EG=BE-BG, \therefore EF=BE-FD.$
6. 如图，在$△ ABC$中，$AB=AC$，$AB＞BC$，点$D$在边$BC$上，$CD=2BD$，点$E$，$F$在线段$AD$上，$∠ 1=∠ 2=∠ BAC$，若$△ BDE$的面积为2，$△ ABC$的面积为21，则$△ CFD$的面积为
9
。答案：6.9 [解析]$\because ∠ 1 = ∠ 2 = ∠ BAC, ∠ 1 = ∠ ABE + ∠ BAE,$
$∠ 2 = ∠ ACF + ∠ FAC, ∠ BAC = ∠ FAC + ∠ BAE,$
$\therefore ∠ ABE = ∠ FAC, ∠ BAE = ∠ ACF.$
又$AB=AC,$
$\therefore △ ABE ≌ △ CAF (\mathrm{ASA}), \therefore S_{△ ABE}=S_{△ CAF}.$
$\because CD=2BD, \therefore S_{△ ACD}=2S_{△ ABD},$
$\therefore S_{△ ACD}=\frac{2}{3}S_{△ ABC}=\frac{2}{3} × 21=14, S_{△ ABD}=\frac{1}{3}S_{△ ABC}=\frac{1}{3} × 21=7,$
$\therefore S_{△ CAF}=S_{△ ABE}=S_{△ ABD}-S_{△ BDE}=7-2=5,$
$\therefore S_{△ CFD}=S_{△ ACD}-S_{△ CAF}=14-5=9.$
$∠ 2 = ∠ ACF + ∠ FAC, ∠ BAC = ∠ FAC + ∠ BAE,$
$\therefore ∠ ABE = ∠ FAC, ∠ BAE = ∠ ACF.$
又$AB=AC,$
$\therefore △ ABE ≌ △ CAF (\mathrm{ASA}), \therefore S_{△ ABE}=S_{△ CAF}.$
$\because CD=2BD, \therefore S_{△ ACD}=2S_{△ ABD},$
$\therefore S_{△ ACD}=\frac{2}{3}S_{△ ABC}=\frac{2}{3} × 21=14, S_{△ ABD}=\frac{1}{3}S_{△ ABC}=\frac{1}{3} × 21=7,$
$\therefore S_{△ CAF}=S_{△ ABE}=S_{△ ABD}-S_{△ BDE}=7-2=5,$
$\therefore S_{△ CFD}=S_{△ ACD}-S_{△ CAF}=14-5=9.$
7.（2026·云南楚雄州期末）如图，在$△ ABC$中，$∠ ACB=90°,AC=BC$，过点$C$在$△ ABC$外作直线$l,AM⊥ l$于点$M,BN⊥ l$于点$N$。试说明：$MN=AM+BN$。
答案：7.$\because ∠ ACB=90°, ∠ ACM + ∠ ACB + ∠ BCN=180°,$
$\therefore ∠ ACM + ∠ BCN=90°.$
$\because AM ⊥ l, BN ⊥ l, \therefore ∠ AMC = ∠ BNC=90°,$
$\therefore ∠ MAC + ∠ ACM=90°, \therefore ∠ MAC = ∠ BCN.$
又$AC=BC, \therefore △ ACM ≌ △ CBN (\mathrm{AAS}),$
$\therefore AM=CN, BN=CM.$
$\because MN=CM+CN, \therefore MN=AM+BN.$
$\therefore ∠ ACM + ∠ BCN=90°.$
$\because AM ⊥ l, BN ⊥ l, \therefore ∠ AMC = ∠ BNC=90°,$
$\therefore ∠ MAC + ∠ ACM=90°, \therefore ∠ MAC = ∠ BCN.$
又$AC=BC, \therefore △ ACM ≌ △ CBN (\mathrm{AAS}),$
$\therefore AM=CN, BN=CM.$
$\because MN=CM+CN, \therefore MN=AM+BN.$
8. 如图,在△ABC中,AD平分∠BAC交BC于点D,M,N分别是AD和AB上的动点,当$S_{△ ABC}=12,AC=8$时,$BM+MN$的最小值等于
3
.答案：
8.3 [解析]如图,在$AC$上取一点$N'$,使$AN'=AN$,连接$MN'$,过点$B$作$BE ⊥ AC$于点$E$.
$\because AD$是$∠ BAC$的平分线,$\therefore ∠ BAD = ∠ CAD.$
$\because AM=AM, \therefore △ ANM ≌ △ AN'M (\mathrm{SAS}),$
$\therefore MN=MN', \therefore BM+MN=BM+MN' \ge BE.$
$\because AC=8, S_{△ ABC}=12, \therefore \frac{1}{2} × 8 · BE=12,$
解得$BE=3, \therefore BM+MN$的最小值是3.
8.3 [解析]如图,在$AC$上取一点$N'$,使$AN'=AN$,连接$MN'$,过点$B$作$BE ⊥ AC$于点$E$.
$\because AD$是$∠ BAC$的平分线,$\therefore ∠ BAD = ∠ CAD.$
$\because AM=AM, \therefore △ ANM ≌ △ AN'M (\mathrm{SAS}),$
$\therefore MN=MN', \therefore BM+MN=BM+MN' \ge BE.$
$\because AC=8, S_{△ ABC}=12, \therefore \frac{1}{2} × 8 · BE=12,$
解得$BE=3, \therefore BM+MN$的最小值是3.
9. 已知:AD是$△ ABC$的角平分线,且$AD ⊥ BC$.
(1)如图(1),求证:$AB=AC$.
(2)如图(2),$∠ ABC=30°$,点E在AD上,连接CE并延长交AB于点F,BG交CA的延长线于点G,且$∠ ABG=∠ ACF$,连接FG.
①求证:$∠ AFG=∠ AFC$;
②若$S_{△ ABG}:S_{△ ACF}=2:3$,且$AG=2$,求AC的长.
(1)如图(1),求证:$AB=AC$.
(2)如图(2),$∠ ABC=30°$,点E在AD上,连接CE并延长交AB于点F,BG交CA的延长线于点G,且$∠ ABG=∠ ACF$,连接FG.
①求证:$∠ AFG=∠ AFC$;
②若$S_{△ ABG}:S_{△ ACF}=2:3$,且$AG=2$,求AC的长.
答案：
9.(1)$\because AD$是$△ ABC$的角平分线，
$\therefore ∠ BAD = ∠ CAD.$
$\because AD ⊥ BC, \therefore ∠ ADB = ∠ ADC,$
在$△ ABD$和$△ ACD$中,$\begin{cases} ∠ BAD=∠ CAD, \\ AD=AD, \\ ∠ ADB=∠ ADC, \end{cases}$
$\therefore △ ABD ≌ △ ACD (\mathrm{ASA}), \therefore AB=AC.$
(2)①$\because AB=AC, ∠ ABC=30°, AD ⊥ BC,$
$\therefore ∠ BAD = ∠ CAD = 60°,$
$\therefore ∠ BAG = 60° = ∠ CAD.$
在$△ BAG$和$△ CAE$中,$\begin{cases} ∠ BAG=∠ CAE, \\ AB=AC, \\ ∠ ABG=∠ ACE, \end{cases}$
$\therefore △ BAG ≌ △ CAE (\mathrm{ASA}), \therefore AG=AE.$
在$△ FAG$和$△ FAE$中,$\begin{cases} AG=AE, \\ ∠ GAF=∠ EAF, \\ AF=AF, \end{cases}$
$\therefore △ FAG ≌ △ FAE (\mathrm{SAS}), \therefore ∠ AFG = ∠ AFC.$
②如图,过点$F$作$FK ⊥ AG$于点$K$.
由①知,$△ BAG ≌ △ CAE. \because S_{△ ABG}:S_{△ ACF}=2:3,$
$\therefore S_{△ CAE}:S_{△ ACF}=2:3, \therefore S_{△ FAE}:S_{△ ACF}=1:3.$
由①知,$△ FAG ≌ △ FAE, \therefore S_{△ FAG}:S_{△ ACF}=1:3,$
$\therefore (\frac{1}{2}AG · FK):(\frac{1}{2}AC · FK)=1:3,$
$\therefore AG:AC=1:3.$
$\because AG=2, \therefore AC=6.$
9.(1)$\because AD$是$△ ABC$的角平分线，
$\therefore ∠ BAD = ∠ CAD.$
$\because AD ⊥ BC, \therefore ∠ ADB = ∠ ADC,$
在$△ ABD$和$△ ACD$中,$\begin{cases} ∠ BAD=∠ CAD, \\ AD=AD, \\ ∠ ADB=∠ ADC, \end{cases}$
$\therefore △ ABD ≌ △ ACD (\mathrm{ASA}), \therefore AB=AC.$
(2)①$\because AB=AC, ∠ ABC=30°, AD ⊥ BC,$
$\therefore ∠ BAD = ∠ CAD = 60°,$
$\therefore ∠ BAG = 60° = ∠ CAD.$
在$△ BAG$和$△ CAE$中,$\begin{cases} ∠ BAG=∠ CAE, \\ AB=AC, \\ ∠ ABG=∠ ACE, \end{cases}$
$\therefore △ BAG ≌ △ CAE (\mathrm{ASA}), \therefore AG=AE.$
在$△ FAG$和$△ FAE$中,$\begin{cases} AG=AE, \\ ∠ GAF=∠ EAF, \\ AF=AF, \end{cases}$
$\therefore △ FAG ≌ △ FAE (\mathrm{SAS}), \therefore ∠ AFG = ∠ AFC.$
②如图,过点$F$作$FK ⊥ AG$于点$K$.
由①知,$△ BAG ≌ △ CAE. \because S_{△ ABG}:S_{△ ACF}=2:3,$
$\therefore S_{△ CAE}:S_{△ ACF}=2:3, \therefore S_{△ FAE}:S_{△ ACF}=1:3.$
由①知,$△ FAG ≌ △ FAE, \therefore S_{△ FAG}:S_{△ ACF}=1:3,$
$\therefore (\frac{1}{2}AG · FK):(\frac{1}{2}AC · FK)=1:3,$
$\therefore AG:AC=1:3.$
$\because AG=2, \therefore AC=6.$