4. As shown in Figure 4, in quadrilateral $ABCD$, $\angle BAD$ and $\angle CAD$ are complementary, $AB=AC$, $AD=5\sqrt{2}$, $BD=8$, $CD=6$. Then the degree measure of $\angle ABD$ is $\qquad$
$4.45^{\circ}$.
As shown in Figure 9, rotate $AD$ clockwise around point $A$ by $90^{\circ}$, and connect $AD^{\prime}$, $BD^{\prime}$, and $DD^{\prime}$. Then
$$
\begin{array}{l}
AD^{\prime}=AD=5 \sqrt{2}, \\
\angle DAD^{\prime}=90^{\circ}.
\end{array}
$$

From $\angle BAD$ and $\angle CAD$ being complementary, we know $\angle BAD^{\prime}=\angle CAD$.
Thus, $\triangle ABD^{\prime} \cong \triangle ACD$
$$
\Rightarrow BD^{\prime}=CD=6.
$$

Therefore, $BD^{2}+BD^{\prime 2}=100=2AD^{2}=D^{\prime}D^{2}$.
So, $\angle DBD^{\prime}=90^{\circ}$.
Hence, points $A$, $D$, $B$, and $D^{\prime}$ are concyclic.
Since $AD=AD^{\prime} \Rightarrow \overparen{AD}=\overparen{AD^{\prime}}$
$$
\Rightarrow \angle ABD=\angle ABD^{\prime}=45^{\circ}.
$$