4. In $\triangle A B C$, $A D$ is the median on side $B C$, $A B=\sqrt{2}$, $A D=\sqrt{6}$, $A C=\sqrt{26}$. Then $\angle A B C=$ $\qquad$
$4.60^{\circ}$.
Extend $B A$ to $E$ such that $A E=A B=\sqrt{2}$, i.e., $B E=2 \sqrt{2}$. Connect $C E$, then $C E / / A D$, and
$$
C E=2 A D=2 \sqrt{6} \text {. }
$$

In $\triangle A C E$, we have $A E^{2}+C E^{2}=2+24=26=A C^{2}$. Therefore, $\angle A E C=90^{\circ}$.
In the right triangle $\triangle B C E$, $C E=\sqrt{3} B E$, so $\angle A B C=60^{\circ}$.