II. (25 points) As shown in Figure 4, given that $E$ is the midpoint of side $AB$ of quadrilateral $ABCD$, $BC=CD=CE$, $\angle B=75^{\circ}$, $\angle D=90^{\circ}$. Find the degree measure of $\angle DAB$.
As shown in Figure 11, draw $E F \perp B C$ at point $F$, and draw $A G \perp B C$ at point $G$. Let $B C = C D = C E = 2 m$.
Since $A G \perp B C$ and $E F \perp B C$,
we have $A G \parallel E F$, and $\angle A G B = 90^{\circ}$.
Also, since $E$ is the midpoint of $A B$, we know that $F$ is the midpoint of $B G$, thus
$$
\begin{array}{l}
A B = 2 B E, A E = B E, \\
B G = 2 B F, A G = 2 E F .
\end{array}
$$

Since $B C = C E$, we have
$$
\angle B E C = \angle B = 75^{\circ},
$$

Therefore, $\angle B C E = 180^{\circ} - \angle B - \angle B E C = 30^{\circ}$.
In the right triangle $\triangle C E F$, we have
$$
\begin{array}{l}
E F = C E \sin \angle B C E = 2 m \sin 30^{\circ} = m, \\
A G = 2 E F = 2 m,
\end{array}
$$

Thus, $A G = C D$.
In the right triangles $\triangle A G C$ and $\triangle A D C$, by the Pythagorean theorem, we get
$$
\begin{array}{l}
A G^{2} + C G^{2} \\
= A C^{2} = A D^{2} + C D^{2} .
\end{array}
$$

Therefore, $A D = G C$, which means quadrilateral $A G C D$ is a parallelogram.
Hence, $\angle D A B = 180^{\circ} - \angle A B C = 105^{\circ}$.