25.(9分)(2024·南通期末)某兴趣小组在学习了三角形相关知识后,对等边三角形进行了再探究.
如图,在等边三角形$ABC$中,过点$B作射线BM// AC$,在射线$CB上取一点P$(不与点$B$,$C$重合),作$\angle APE = 60^{\circ}$,$\angle APE的边PE交射线BM于点E$.
(1)【动手操作】
如图①,若点$P在线段CB$上,图中与$\angle EPB$相等的角为______；
(2)【问题探究】
在(1)的基础上,探究线段$PA与PE$的数量关系,并说明理由；
(3)【拓展延伸】
当点$P在射线CB$上移动时,用等式表示线段$BC$,$BP$,$BE$之间的数量关系,并说明理由.

(1)∠PAC
(2)PA = PE.理由如下:如图①,延长EB至点H,使BH = BP,连接PH.∵△ABC是等边三角形,∴AB = BC,∠ACB = ∠ABC = 60°.∵BM//AC,∴∠ACB = ∠CBH = 60°.又∵BP = BH,∴△BPH是等边三角形,∴PH = BP = BH,∠H = 60° = ∠ABC = ∠APE = ∠BPH,∴∠APB = ∠EPH,∴△APB≌△EPH(ASA),∴PA = PE.
　
(3)当点P在BC上时,BC = BP + BE;当点P在线段CB的延长线上时,BE = BP + BC.理由如下:当点P在BC上时,由(2)可知,△APB≌△EPH,∴AB = EH,∴BC = EH = EB + BH = BE + BP.当点P在线段CB的延长线上时,如图②,在BE上截取BH = BP,连接PH.∵△ABC是等边三角形,∴AB = BC,∠ACB = ∠ABC = 60°.∵BM//AC,∴∠ACB = ∠PBH = 60°.又∵BP = BH,∴△BPH是等边三角形,∴PH = BP = BH,∠BHP = 60° = ∠ABC = ∠APE = ∠BPH,∴∠APB = ∠EPH,∠EHP = ∠ABP = 120°,∴△APB≌△EPH(ASA),∴EH = AB,∴BE = BH + EH = BP + BC.

26.(11分)如图①,$\triangle ABC$中,$CD\perp AB于点D$,

且$BD:AD:CD = 2:3:4$.
(1)试说明$\triangle ABC$是等腰三角形.
(2)已知$S_{\triangle ABC} = 40\mathrm{cm}^2$,如图②,动点$M从点B$出发,以每秒1 cm的速度沿线段$BA向点A$运动,同时动点$N从点A$出发,以相同速度沿线段$AC向点C$运动,当其中一点到达终点时整个运动都停止.设点$M运动的时间为t$ s.
①若$\triangle DMN的边与BC$平行,求$t$的值.
②若点$E是边AC$的中点,问：在点$M$运动的过程中,$\triangle MDE$能否成为等腰三角形？若能,求出$t$的值；若不能,请说明理由.

(1)设BD = 2x,AD = 3x,CD = 4x,则AB = 5x,在Rt△ACD中,$AC^{2}=AD^{2}+CD^{2}$,∴AC = 5x,∴AB = AC,∴△ABC是等腰三角形.
(2)$S_{△ABC}=\frac{1}{2}×5x×4x = 40$,而x ＞ 0,∴x = 2,则BD = 4cm,AD = 6cm,CD = 8cm,AC = 10cm.
①当MN//BC时,AM = AN,即10 - t = t,∴t = 5;当DN//BC时,AD = AN,得t = 6.∴若△DMN的边与BC平行,则t的值为5或6.
②能.当点M在BD上,即0 ≤ t ＜ 4时,△MDE为钝角三角形,但DM ≠ DE;当t = 4时,点M运动到点D,不构成三角形;当点M在DA上,即4 ＜ t ≤ 10时,△MDE为等腰三角形,有3种可能.若DE = DM,则t - 4 = 5,∴t = 9;若ED = EM,则点M运动到点A,∴t = 10;若MD = ME = t - 4.过点E作EF⊥AB于点F,∵ED = EA,∴DF = AF =$\frac{1}{2}AD = 3$cm,在Rt△AEF中,EF = 4cm.∵BM = tcm,BF = 7cm,∴FM = |t - 7|cm.在Rt△EFM中,$(t - 4)^{2}-(t - 7)^{2}=4^{2}$,∴$t = \frac{49}{6}$.综上所述,符合要求的t的值为9或10或$\frac{49}{6}$.