如图, $A C$ 平分 $\angle B A D$, 过点 $C$ 作 $C E \perp A B$ 于 $E$, 并且 $2 A E=A B+A D$, 则下列结论, 正确的是 \$ \qquad \$ . (填序号)

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(1) $A B=A D+2 B E$; (2) $\angle D A B+\angle D C B=180^{\circ}$; (3) $C D=C B$; (4) $S_{\triangle A B C}=S_{\triangle A C D}+S_{\triangle B C E}$
解：如图, 过 $C$ 作 $C F \perp A D$ 于 $F$,

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$\because A C$ 平分 $\angle B A D, C E \perp A B, C F \perp A D$,

$\therefore C F=C E$,

$\therefore R t \triangle A C F \cong R t \triangle A C E(H L)$,

$\therefore A F=A E$,

$\therefore A B+A D=(A E+B E)+(A F-D F)=2 A E+B E-D F$,

又 $\because A B+A D=2 A E$,

$\therefore B E=D F$,

$\therefore A B-A D=(A E+B E)-(A F-D F)=B E+D F=2 B E$,

即 $A B=A D+2 B E$, 故(1)正确;

$\because B E=D F, \angle C E B=\angle F=90^{\circ}, C F=C E$,

$\therefore \triangle C D F \cong \triangle C B E(S A S)$,

$\therefore \angle B=\angle C D F, C D=C B$, 故(3)正确;
又 $\because \angle A D C+\angle C D F=180^{\circ}$,

$\therefore \angle A D C+\angle B=180^{\circ}$,

$\therefore$ 四边形 $A B C D$ 中, $\angle D A B+\angle B C D=360^{\circ}-180^{\circ}=180^{\circ}$, 故(2)正确;

$\because A B=A D+2 B E, C E=C F$,

$\therefore$ 由等式性质可得 ${ }_{2}^{1} A B \times C E=\frac{1}{2} A D \times C F+2 \times \frac{1}{2} B E \times C E$,

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故答案为(1)(2)(3).