12. (江西中考)如图,CA平分$∠DCB,CB= CD$,DA的延长线交BC于点E,若$∠EAC= 49^{\circ}$,则$∠BAE$的度数为____.

82°

13. 如图,在$△ABC$中,E是BC上的一点,$EC= 2BE$,点D是AC的中点,设$△ABC,△ADF,△BEF的面积分别为S_{△ABC},S_{△ADF},S_{△BEF}$,且$S_{△ABC}= 12$,则$S_{△ADF}-S_{△BEF}= $____.

2

14. 如图,在长方形ABCD中,$AB= 4,AD= 6$.延长BC到点E,使$CE= 2$,连接DE,动点P从点B出发,以每秒2个单位长度的速度沿$BC-CD-DA$向终点A运动,设点P的运动时间为t秒,当$△ABP和△DCE$全等时,t的值为____.

1或7 解析:因为AB = CD,若∠ABP = ∠DCE = 90°,BP = CE = 2,根据SAS证得△ABP≌△DCE,由题意得BP = 2t = 2,所以t = 1.因为AB = CD,若∠BAP = ∠DCE = 90°,AP = CE = 2,根据SAS证得△BAP≌△DCE,由题意得AP = 16 - 2t = 2,解得t = 7.所以当t的值为1或7时,△ABP和△DCE全等.

15. 如图,在$△ABC$中,$AB= AC$,D,E是$△ABC$内的两点,AE平分$∠BAC,∠D= ∠DBC= 60^{\circ}$,若$BD= 6cm,DE= 4cm$,则BC的长是____cm.

10 解析:如图,延长DE交BC于点M,延长AE交BC于点N.∵AB = AC,AE平分∠BAC,∴AN⊥BC,BN = CN.∵∠DBC = ∠D = 60°,∴△BDM为等边三角形,∴BD = DM = BM = 6 cm.∵DE = 4 cm,∴EM = 6 - 4 = 2(cm).∵△BDM为等边三角形,∴∠DMB = 60°.∵AN⊥BC,∴∠ENM = 90°,∴∠NEM = 30°,∴NM = $\frac{1}{2}$ME = 1 cm,∴BN = 6 - 1 = 5(cm),∴BC = 2BN = 10 cm.故答案为10.

16. 如图,边长为6的等边$△ABC$,F是边AC的中点,点D是线段BF上的动点,连接AD,在AD的右侧作等边$△ADE$,连接CD,CE,EF,则下列说法正确的是____(填序号).
①$BF⊥AC$;②$∠DEC= ∠DCE$;③$AE= CD$;④$△ADE$的周长最小值为9;⑤当$△AEF$周长最小时,$∠AFE= 60^{\circ}$;⑥$∠ACE$的大小随着点D的移动而变化.

①②③④ 解析:∵△ABC是等边三角形,F是边AC的中点,∴BF⊥AC.故①正确;∵BF是线段AC的垂直平分线,∴AD = CD.∵△ADE是等边三角形,∴AD = ED = AE,∴AE = ED = CD.故③正确;∴∠DEC = ∠DCE.故②正确;∵点D在线段BF上,∴当AD⊥BF,即点D与点F重合时,AD最小,即此时△ADE的周长最小.∵等边三角形△ABC的边长为6,F是边AC的中点,∴AD最小 = AF = 3,∴△ADE的周长的最小值为AD + DE + AE = 3AD = 9.故④正确;∵△ABC,△ADE都是等边三角形,∴AB = AC,AD = AE,∠BAC = ∠DAE = 60°,∴∠BAD = ∠CAE.∴△BAD≌△CAE(SAS).∴∠ACE = ∠ABD = 30°.故⑥错误;∴∠BCE = ∠BCA + ∠ACE = 90°,即点E在射线CE(射线CE⊥BC)上运动.如图所示,作点A关于射线CE的对称点M,连接ME,MC,∴AE = ME,∴△AEF的周长 = AF + AE + EF = AF + EM + EF,∴当E,F,M三点共线,即点E与点E'重合时,EF + EM最小,即△AEF的周长最小.∵点A与点M关于射线CE对称,∴AC = MC,∠ACE = ∠MCE = 30°,∴∠ACM = 60°.∴△ACM是等边三角形.又∵F是边AC的中点,∴AF⊥MF,∴∠AFE' = 90°.故⑤错误.正确的是①②③④.

17. (7分)新趋势 过程性学习 (2024·淄博中考)如图,已知$AB= CD$,点E,F在线段BD上,且$AF= CE$.
请从①$BF= DE$;②$∠BAF= ∠DCE$;③$AF= CF$中.选择一个合适的选项作为已知条件,使得$△ABF\cong △CDE$.
你添加的条件是:____(只填写一个序号).
添加条件后,请证明$AE// CF$.

①或②(只选一个即可)
当选取①时,在△ABF与△CDE中,{AB = CD,AF = CE,BF = DE,∴△ABF≌△CDE(SSS),∴∠B = ∠D.∵BF = DE,∴BF + EF = DE + EF,∴BE = DF.在△ABE与△CDF中,{AB = CD,∠B = ∠D,BE = DF,∴△ABE≌△CDF(SAS),∴∠AEB = ∠CFD,∴AE//CF.
当选取②时,在△ABF与△CDE中,{AB = CD,∠BAF = ∠DCE,AF = CE,∴△ABF≌△CDE(SAS),∴∠B = ∠D,BF = DE,∴BF + EF = DE + EF,∴BE = DF.在△ABE与△CDF中,{AB = CD,∠B = ∠D,BE = DF,∴△ABE≌△CDF(SAS),∴∠AEB = ∠CFD,∴AE//CF.

18. (8分)如图,在等腰三角形ABC中,$AB= AC$,点D,E分别在边AB,AC上,且$AD= AE$,连接BE,CD交于点F.
(1)判断$∠ABE与∠ACD$的数量关系,并说明理由;
(2)求证:过点A,F的直线垂直平分线段BC.

(1)∠ABE = ∠ACD.理由:在△ABE和△ACD中,∵AB = AC,∠A = ∠A,AE = AD,∴△ABE≌△ACD(SAS),∴∠ABE = ∠ACD.
(2)连接AF.∵AB = AC,∴∠ABC = ∠ACB.由(1)可知∠ABE = ∠ACD,∴∠FBC = ∠FCB,∴FB = FC.∵AB = AC,∴点A,F均在线段BC的垂直平分线上,即过点A,F的直线垂直平分线段BC.