1. 如图，在$△ ABC$中，$AD ⊥ BC$于点$D$，$∠ B=2∠ C$. 求证：$AB+BD=CD$.

1. 如图,在 CD 上取一点 E,使 DE = BD,连结 AE.
∵ AD ⊥ BC,
∴ △ABE 是等腰三角形,
∴ AB = AE, ∠B = ∠AEB.
∵ ∠B = ∠AEB = 2∠C, ∠AEB = ∠C + ∠EAC,
∴ ∠EAC = ∠C,
∴ AE = EC,
∴ AB+BD = CE+ED = CD.

2. ★★★ 如图,在$△ ABC$中,D是BC边上的一点,E是AD边上的一点,$AD=DC$,$∠ DEC=∠ ABC$. 求证:$AB=CE$.

2. 如图,延长 CB 至点 H,使 AH = AC,则 ∠AHB = ∠ACD.
∵ AD=CD,
∴ ∠DAC = ∠ACD,
∴ ∠AHB = ∠DAC.
∵ ∠DEC = ∠ABC,
∴ ∠ABH = ∠CEA. 又
∵ AH = AC,
∴ △ABH ≅ △CEA(AAS),
∴ AB=CE.

3. 如图,在△ABC中,AB=AC,∠A=100°,∠ABC的平分线交AC于点E.求证:BE+AE=BC.

3. 如图,在 BC 上分别截取点 D,F,使BD=AB,BF=BE.
∵ BE 是∠ABC 的平分线,
∴ ∠ABE = ∠DBE.
∵ BD = AB,BE = BE,
∴ △ABE≅△DBE(SAS),
∴ ∠BDE = ∠A = 100°,AE = DE.
∵ AB = AC,∠A = 100°,
∴ ∠C = ∠ABC = 40°.
∵ BE 是∠ABC 的平分线,
∴ ∠ABE = ∠DBE = 20°.
∵ BE = BF,
∴ ∠BFE = ∠BEF = 80°.
∵ ∠EDF = 180°−∠BDE = 80°,
∴ ∠EDF = ∠BFE,
∴ DE=EF.
∵ ∠BFE = ∠C + ∠FEC,
∴ ∠FEC = ∠BFE−∠C = 40°,
∴ ∠FEC = ∠C,
∴ EF = FC,
∴ AE = FC.
∵ BF+FC = BC,
∴ BE+AE = BC.