Example 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 $\angle DAB$. ${ }^{[4]}$
Solve As shown in Figure 4, draw $E F \perp B C$ at point $F$, and draw $A G \perp B C$ at point $G$.
Let $B C=C D$
$$
=C E=2 a \text {. }
$$

Since $A G \perp B C, E F \perp$ $B C, E$ is the midpoint of $A B$, we know that $F$ is the midpoint of $B G$.
Thus, $A B=2 B E, A E=B E$,
$$
B G=2 B F, A G=2 E F \text {. }
$$

Also, since $B C=C E$, we know $\angle B E C=\angle B=75^{\circ}$. 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
$E F=C E \sin \angle B C E=a, A G=2 E F=2 a$, so $A G=C E=C D$.
In the right triangles $\triangle A G C$ and $\triangle A D C$, by the Pythagorean theorem we get
$$
A G^{2}+C G^{2}=A C^{2}=A D^{2}+C D^{2} .
$$

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