4. As shown in Figure 2, in the right trapezoid $A B C D$, $A D / /$ $B C, \angle A B C=90^{\circ}, A B=B C, E$ is a point on side $A B$, $\angle B C E=15^{\circ}$, and $A E=A D$. Connect $D E$ and diagonal $A C$ intersect at point $H$, and connect $B H$. Given the following conclusions:
(1) $\triangle A C D \cong \triangle A C E$;
(2) $\triangle C D E$ is an equilateral triangle;
(3) $\frac{E H}{B E}=2$;
(4) $\frac{S_{\triangle E B C}}{S_{\triangle E H C}}=\frac{A H}{C H}$.

Among them, the correct conclusions are ( ).
(A)(1)(2)
(B)(1)(2)(4).
(C)(3)(4)
(D)(1)(2)(3)(4)
4. B.

It is known that $\triangle A B C$ and $\triangle A D E$ are both isosceles right triangles.
Therefore, $\angle B A C=45^{\circ}$, which means $A H$ bisects $\angle E A D$.
Thus, $\triangle A C D \cong \triangle A C E$, so conclusion (1) is correct.
Since $\angle A C B=45^{\circ}$ and $\angle B C E=15^{\circ}$, we have
$\angle D C A=\angle E C A=45^{\circ}-15^{\circ}=30^{\circ}$.
Thus, $\angle E C D=2 \angle D C A=60^{\circ}$.
Since $C E=C D$, $\triangle C D E$ is an equilateral triangle, so conclusion (2) is correct.
Let the side length of $\triangle C D E$ be $2 a$. Then
$$
\begin{array}{l}
E H=A H=a, C H=\sqrt{3} a, A E=\sqrt{2} a, \\
A B=\frac{1}{\sqrt{2}} A C=\frac{A H+C H}{\sqrt{2}}=\frac{\sqrt{6}+\sqrt{2}}{2} a, \\
B E=A B-A E=\frac{\sqrt{6}-\sqrt{2}}{2} a .
\end{array}
$$

Therefore, $\frac{E H}{B E}=\frac{\sqrt{6}+\sqrt{2}}{2}$, so conclusion (3) is incorrect.
Also, $S_{\triangle E B C}=\frac{1}{2} B C \cdot B E=\frac{1}{2} a^{2}$,
$S_{\triangle E H C}=\frac{1}{2} S_{\triangle C D E}=\frac{\sqrt{3}}{2} a^{2}$,
then $\frac{S_{\triangle E B C}}{S_{\triangle E H C}}=\frac{\sqrt{3}}{3}=\frac{A H}{C H}$, so conclusion (4) is correct.