1. (原创题)如图,在四边形 $ABCD$ 中,$AC$ 平分 $∠BAD$,$∠B + ∠ADC = 180^\circ$,$CE⊥AB$ 于点 $E$.
(1)求证:$CB = CD$；
(2)若 $AB + AD = 10$,求 $AE$ 的长.

解: (1) 过点 $ C $ 作 $ CF \perp AD $ 于点 $ F $.
$ \because AC $ 平分 $ \angle BAD $, $ CE \perp AB $,
$ \therefore CE = CF $.
$ \because \angle B + \angle ADC = 180 ^ { \circ } $,
$ \angle CDF + \angle ADC = 180 ^ { \circ } $,
$ \therefore \angle CDF = \angle B $,
$ \therefore \triangle CFD \cong \triangle CEB $,
$ \therefore CB = CD $;
(2) $ \because \angle FAC = \angle BAC $,
$ \angle F = \angle CEA = 90 ^ { \circ } $, $ AC = AC $,
$ \therefore \triangle CAF \cong \triangle CAE $,
$ \therefore AE = AF $.
又 $ \because \triangle CDF \cong \triangle CBE $,
$ \therefore BE = DF $,
$ \begin{align}\therefore AB + AD & = ( AE + BE ) + ( AF - DF ) \\& = AE + AF \\& = 2 AE \\& = 10,\end{align} $
$ \therefore AE = 5 $.

2. (原创题)如图,在四边形 $ABCD$ 中,$AC$ 平分 $∠BAD$,$CD = CB$,$CE⊥AB$ 于点 $E$.
(1)求证:$∠ADC + ∠ABC = 180^\circ$；

(2)若 $BE = 2$,求 $AB - AD$ 的值.

解: (1) 过点 $ C $ 作 $ CF \perp AD $ 于点 $ F $.
$ \because AC $ 平分 $ \angle BAD $, $ CE \perp AB $,
$ \therefore CE = CF $.
$ \because CD = CB $,
$ \therefore \mathrm { Rt } \triangle CFD \cong \mathrm { Rt } \triangle CEB $,
$ \therefore \angle CDF = \angle CBE $.
$ \because \angle CDF + \angle ADC = 180 ^ { \circ } $,
$ \therefore \angle ADC + \angle ABC = 180 ^ { \circ } $;
(2) 在 $ \mathrm { Rt } \triangle CAF $ 和 $ \mathrm { Rt } \triangle CAE $ 中,
$ \because CF = CE $, $ AC = AC $,
$ \therefore \mathrm { Rt } \triangle CAF \cong \mathrm { Rt } \triangle CAE ( \mathrm { HL } ) $,
$ \therefore AE = AF $.
又 $ \because \triangle CFD \cong \triangle CEB $,
$ \therefore DF = BE $,
$ \begin{align}\therefore AB - AD & = ( AE + BE ) - ( AF - DF ) \\& = BE + DF \\& = 2 BE \\& = 4.\end{align} $

3. (原创题)如图,在四边形 $ABDC$ 中,$AB = AC$,$∠BAC + ∠BDC = 180^\circ$,过点 $A$ 作 $AE⊥CD$,垂足为 $E$,$DE - EC = 1$.求 $BD$ 的长.

解: 过点 $ A $ 作 $ AF \perp BD $,
交 $ DB $ 的延长线于点 $ F $, 连接 $ AD $.
可得 $ \angle BAC + \angle ABD + \angle BDC + \angle C = 180 ^ { \circ } \times 2 = 360 ^ { \circ } $.
$ \because \angle BAC + \angle BDC = 180 ^ { \circ } $,
$ \therefore \angle ABD + \angle C = 180 ^ { \circ } $.
又 $ \because \angle ABD + \angle ABF = 180 ^ { \circ } $,
$ \therefore \angle C = \angle ABF $.
$ \because AF \perp BD $, $ AE \perp CD $,
$ \therefore \angle F = \angle AEC = 90 ^ { \circ } $.
$ \because AB = AC $,
$ \therefore \triangle ABF \cong \triangle ACE $,
$ \therefore BF = EC $, $ AF = AE $.
$ \because AD = AD $,
$ \therefore \mathrm { Rt } \triangle ADF \cong \mathrm { Rt } \triangle ADE $,
$ \therefore DF = DE $,
$ \therefore BD = DF - BF = DE - EC = 1 $.