8. (2025·太原模拟)如图,已知$△ABC$,点D是AC边上一点,根据尺规作图的痕迹,能确定线段BD是$△ABC$的()

A. 中线
B. 角平分线
C. 高线
D. 以上均正确

B 解析:如图,根据作图可知,BE = BF,BG = BH,∵∠EBG = ∠FBH,∴△BEG≌△BFH(SAS),∴∠EHM = ∠FGM.∵BH - BE = BG - BF,∴EH = FG.∵∠EMH = ∠FMG,∴△EMH≌△FMG(AAS),∴GM = HM.∵BG = BH,BM = BM,∴△BMH≌△BMG(SSS),∴∠HBM = ∠GBM,∴BD平分∠ABC,即线段BD是△ABC的角平分线.故选B.

9. (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$24cm^{2}$,则$S_{△QPO}-S_{△CDO}= $____$cm^{2}$.

24 解析:∵AB//DC,∴∠Q = ∠DCM,∠P = ∠CDN.∵M和N分别是AD和BC的中点,∴AM = DM,BN = CN.∵∠AMQ = ∠DMC,∴△AQM≌△DCM(AAS),同理可得△BPN≌△CDN,∴S△AQM = S△DCM,S△BPN = S△CDN,∴S△QPO = S△AQM + S△BPN + S五边形ABNOM = S△CDM + S△CDN + S五边形ABNOM = S梯形ABCD + S△CDO,∴S△QPO - S△CDO = S梯形ABCD = 24cm².

10. (2025·重庆期末)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)$△AEF\cong △DHF$;
(2)$∠B= 2∠GDC$.

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

11. (2024·宁波期末)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,$\frac {AP}{AQ}$的值为____.

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,∵AB = AC,AP + AQ = AB,AB = AP + BP,AC = AQ + CQ,∴AQ = BP,CQ = AP = BH.∵AC//BE,∴∠A = ∠EBH.在△ACP和△BEH中,$\left\{\begin{array}{l} AC = BE,\\ ∠A = ∠EBH,\\ AP = BH,\end{array}\right. $∴△ACP≌△BEH(SAS),∴CP = HE.∵AB = AC,∴∠ACB = ∠ABC.在△CBQ和△BCH中,$\left\{\begin{array}{l} CB = BC,\\ ∠BCQ = ∠CBH,\\ CQ = BH,\end{array}\right. $∴△CBQ≌△BCH(SAS),∴CH = BQ,∴BQ + CP = CH + HE,∴当C,E,H三点共线时,BQ + CP有最小值,此时,∵AC//BE,∴∠A = ∠EBA,∠ACH = ∠BEH.又∵AC = BE,∴△ACH≌△BEH(ASA),∴AH = BH,∴点H是AB的中点,∴AP = BH = $\frac{1}{2}$AB,∴点P与点H重合,∴BP = BH = AQ = AP,∴$\frac{AP}{AQ}$ = 1.

12. (2025·无锡校级月考)已知$△ABC$中,$∠ACB= 90^{\circ },AC= CB$,D为直线BC上一动点,连接AD,在直线AC右侧作$AE⊥AD$,且$AE= AD$.
(1)如图①,当点D在线段BC上时,过点E作$EH⊥AC$于H,求证:$EH= AC$;
(2)如图②,当点D在线段BC的延长线上时,连接BE交CA的延长线于点M,求证:$S_{△ABM}= S_{△AME}$;
(3)当点D在直线CB上时,连接BE交直线AC于M,若$AC= 3CM$,则$\frac {S_{△ADB}}{S_{△AEM}}= $____.

(1)∵AE⊥AD,EH⊥AC,∴∠EHA = ∠EAD = ∠ACB = 90°,∴∠DAC + ∠ADC = 90°,∠DAC + ∠EAH = 90°,∴∠ADC = ∠EAH.在△ADC和△EAH中,$\left\{\begin{array}{l} ∠ACD = ∠EHA,\\ ∠ADC = ∠EAH,\\ AD = EA,\end{array}\right. $∴△ADC≌△EAH(AAS),∴EH = AC.
(2)如图①,过点E作EN⊥AM,交AM的延长线于点N,∵AE⊥AD,EN⊥AM,∴∠EAD = ∠ANE = ∠ACB = 90°,∴∠ACD = 180° - ∠ACB = 180° - 90° = 90°,∴∠ACD = ∠ENA,∠DAC + ∠ADC = 90°.∵∠DAC + ∠EAN = 180° - ∠EAD = 180° - 90° = 90°,∴∠ADC = ∠EAN.在△ADC和△EAN中,$\left\{\begin{array}{l} ∠ACD = ∠ENA,\\ ∠ADC = ∠EAN,\\ AD = EA,\end{array}\right. $∴△ADC≌△EAN(AAS),∴EN = AC.∵AC = BC,∴EN = BC.∵S△ABM = $\frac{1}{2}$·AM·BC,S△AME = $\frac{1}{2}$·AM·EN,∴S△ABM = S△AME.


(3)1或$\frac{1}{2}$ 解析:分三种情况:
①当点D在线段BC上时,如图②,过点E作EH⊥AC于H,∵AC = 3CM,∴设CM = a,则AC = 3a,由(1)得△ADC≌△EAH,∴AH = DC,EH = AC = 3a.∵BC = AC = 3a,∴BC = EH = 3a.∵EH⊥AC,∴∠EHM = ∠BCM = 90°.在△BCM和△EHM中,$\left\{\begin{array}{l} ∠BMC = ∠EMH,\\ ∠BCM = ∠EHM,\\ BC = EH,\end{array}\right. $∴△BCM≌△EHM(AAS),∴HM = CM = a,∴AH = AC - CM - HM = 3a - a - a = a,∴AM = AH + HM = a + a = 2a.∵CD = AH,∴BD = BC - CD = BC - AH = 3a - a = 2a,∴$\frac{S_{△ADB}}{S_{△AEM}} = \frac{\frac{1}{2} \cdot BD \cdot AC}{\frac{1}{2} \cdot AM \cdot EH} = \frac{\frac{1}{2} \cdot 2a \cdot 3a}{\frac{1}{2} \cdot 2a \cdot 3a} = 1$.
②当点D在线段BC的延长线上时,如图③,由图可得AC ＜ CM,∴AC = 3CM不可能,∴此种情况不存在.


③当点D在CB的延长线上时,如图④,过点E作EN⊥AM,交AM的延长线于点N,∵AC = 3CM,∴设CM = a,则AC = 3a.∵AE⊥AD,EN⊥AM,∴∠EAD = ∠ENA = ∠ACB = 90°,∴∠DAC + ∠ADC = 90°,∠DAC + ∠EAN = 90°,∴∠ADC = ∠EAN.在△ADC和△EAN中,$\left\{\begin{array}{l} ∠ACD = ∠ENA,\\ ∠ADC = ∠EAN,\\ AD = EA,\end{array}\right. $∴△ADC≌△EAN(AAS),∴EN = AC,AN = DC.∵AC = BC,∴EN = BC.∵∠ACB = 90°,∴∠BCM = 180° - ∠ACB = 180° - 90° = 90°.在△BCM和△ENM中,$\left\{\begin{array}{l} ∠BMC = ∠EMN,\\ ∠BCM = ∠ENM,\\ BC = EN,\end{array}\right. $∴△BCM≌△ENM(AAS),∴NM = CM = a,BC = EN = AC = 3a,∴AN = AC + CM + NM = 3a + a + a = 5a,AM = AC + CM = 3a + a = 4a.∵DC = AN,∴BD = DC - BC = AN - BC = 5a - 3a = 2a,∴$\frac{S_{△ADB}}{S_{△AEM}} = \frac{\frac{1}{2} \cdot BD \cdot AC}{\frac{1}{2} \cdot AM \cdot EN} = \frac{\frac{1}{2} \cdot 2a \cdot 3a}{\frac{1}{2} \cdot 4a \cdot 3a} = \frac{1}{2}$.
综上所述，$\frac{S_{△ADB}}{S_{△AEM}} = 1$或$\frac{1}{2}$.