1. 如图,在$△ABC$中,$AB= AC,∠BAC= 90^{\circ }$,CD平分$∠ACB$交AB于点D.求证:$BC= AC+AD$.

证明:过点 D 作 $ DE \perp BC $ 于点 E.
$ \because AC = AB, \angle BAC = 90^\circ $,
$ \therefore \angle B = \angle ACB = 45^\circ $.
$ \because \angle BED = 90^\circ $,
$ \therefore \angle B = \angle BDE = 45^\circ $,
$ \therefore BE = DE $.
$ \because CD $ 平分 $ \angle ACB $,
$ \therefore \angle ACD = \angle ECD $.
$ \because CD = CD $,
$ \angle BAC = \angle DEC = 90^\circ $,
$ \therefore \triangle ACD \cong \triangle ECD $,
$ \therefore AC = EC, AD = DE = BE $.
$ \because BC = EC + BE $,
$ \therefore BC = AC + AD $.

2. 如图,在$△ABC$中,$CA= CB,∠ACB= 108^{\circ }$,BD平分$∠ABC$交AC于点D.求证:$AB= AD+BC$.

证明:方法一:(截长法)在 AB 上截取 $ BE = BC $,连接 DE,
$ \because BD $ 平分 $ \angle ABC $,
$ \therefore \angle 1 = \angle 2 $.
又 $ \because BD = BD $,
$ \therefore \triangle BCD \cong \triangle BED $,
$ \therefore \angle BED = \angle ACB = 108^\circ $,
$ \therefore \angle AED = 72^\circ $.
$ \because CA = CB, \angle ACB = 108^\circ $,
$ \therefore \angle A = \angle ABC = 36^\circ $,
$ \therefore \angle ADE = 72^\circ $,
$ \therefore \angle AED = \angle ADE $,
$ \therefore AD = AE $,
$ \therefore AB = BE + AE = BC + AD $;
方法二:(补短法)如图,延长 BC 至点 F,使 $ BF = AB $,连接 FD,只需证 $ AD = DF = CF $ 即可.

3. (原创题)如图,在四边形ABCD中,BD平分$∠ABC,∠A= α$(α为钝角),$∠C= 2α-180^{\circ }$.求证:$BC= AB+CD$.

证明:在 BC 上截取 $ BE = BA $,连接 DE.
$ \because BD $ 平分 $ \angle ABC $,
$ \therefore \angle ABD = \angle DBE $.
$ \because BD = BD $,
$ \therefore \triangle ABD \cong \triangle EBD(SAS) $,
$ \therefore \angle DEB = \angle A = \alpha $,
$ \therefore \angle DEC = 180^\circ - \angle DEB $
$ = 180^\circ - \alpha $,
$ \therefore \angle EDC = 180^\circ - \angle DEC - \angle C $
$ = 180^\circ - (180^\circ - \alpha) - (2\alpha - 180^\circ) $
$ = 180^\circ - \alpha $,
$ \therefore \angle EDC = \angle DEC $,
$ \therefore CE = CD $,
$ \therefore BC = BE + CE = AB + CD $.