1. (1)如图,已知$AB= AC,∠BAC= 90^{\circ },∠MAN= 45^{\circ }$,过点 C 作$NC⊥AC$交 AN 于点 N,过点 B 作$BM⊥AB$交 AM 于点 M,连接 MN,当$∠MAN在∠BAC$内部时,求证:$BM+CN= MN;$

(2)如图,在(1)的条件下,当 AM 和 AN 在 AB 两侧时,(1)的结论是否成立?请说明理由.

(1)证明:延长 MB 至点 G,使 $ BG = CN $,连接 AG,证 $ \triangle ABG \cong \triangle ACN(SAS) $,$ \therefore AN = AG $,$ \angle BAG = \angle NAC $.$ \because \angle GAM = \angle GAB + \angle BAM $$ = \angle CAN + \angle BAM $$ = 45^{\circ} $$ = \angle MAN $,证 $ \triangle AMN \cong \triangle AMG(SAS) $,$ \therefore MN = MG = BM + BG = BM + NC $;(2)解:不成立,结论是 $ MN = CN - BM $.理由如下:在 CN 上截取 CG,使 $ CG = BM $,连接 AG,证 $ \triangle ACG \cong \triangle ABM(SAS) $,$ \therefore AG = AM, \angle CAG = \angle BAM $,$ \therefore \angle MAG = \angle BAC = 90^{\circ} $.$ \because \angle MAN = 45^{\circ} $,$ \therefore \angle GAN = 45^{\circ} = \angle MAN $,证 $ \triangle MAN \cong \triangle GAN(SAS) $,$ \therefore MN = GN $.$ \because NG = CN - CG $,$ \therefore MN = CN - BM $.

2. 如图,在$△ABC$中,$CA= CB,∠ACB= 120^{\circ }$,E 为 AB 上一点,$∠DCE= 60^{\circ },∠DAE= 120^{\circ }$.求证:$DE-AD= BE.$

证明:(补短法)延长 EB 至点 F,使 $ BF = AD $,连接 CF,可求 $ \angle CAB = \angle CBA = 30^{\circ} $,$ \therefore \angle DAC = \angle CBF = 150^{\circ} $.又 $ \because CA = CB $,$ \therefore \triangle CBF \cong \triangle CAD $,$ \therefore CD = CF $.$ \because \angle DCE = \angle FCE = 60^{\circ}, CE = CE $,$ \therefore \triangle CED \cong \triangle CEF $,$ \therefore DE = EF $,$ \therefore DE - AD = EF - BF $$ = BE $.

3. 如图,在$△ABC$中,$CA= CB,∠ACB= 120^{\circ }$,E 为 AB 上一点,点 D 在$△ABC$的外部,且$∠DCE= ∠DAE= 60^{\circ }$.求证:$AD+DE= BE.$

证明:(截长法)在 BE 上截取 $ BF = AD $,连接 CF.$ \because \angle ACB = 120^{\circ}, CA = CB $,$ \therefore \angle CAB = \angle B = 30^{\circ} $,$ \therefore \angle DAC = 30^{\circ} = \angle B $,$ \because AD = BF, CA = CB $,$ \therefore \triangle CBF \cong \triangle CAD(SAS) $,$ \therefore CD = CF, \angle BCF = \angle ACD $,$ \therefore \angle DCF = \angle ACB = 120^{\circ} $,$ \therefore \angle DCE = \angle ECF = 60^{\circ} $.又 $ \because CE = CE $,$ \therefore \triangle CED \cong \triangle CEF(SAS) $,$ \therefore DE = EF $,$ \therefore AD + DE = BF + EF = BE $.