(4) In tetrahedron $ABCD$, $CD \perp BC, AB \perp BC, CD=AC, AB=$ $BC=1$ plane $BCD$ and plane $ABC$ form a $45^{\circ}$ dihedral angle, then the distance from point $B$ to plane $ACD$ is $\qquad$.
$4 \frac{\sqrt{3}}{3}$ Hint: $D C=A C=\sqrt{2}$, draw $D E \perp$ plane $A B C$, the foot of the perpendicular is $E$, connect $C E, A E$, by the inverse of the three perpendiculars theorem, $E C \perp B C$, so $\angle D C E=45^{\circ}$, hence $C E=D E=\frac{\sqrt{2}}{2} D C=1, V_{A B C D}=\frac{1}{3} D E \cdot S_{A B C}=\frac{1}{6}$. Since $A B C E$ is a square, $A E=1$, then $A D=\sqrt{2}$, thus the area of the equilateral triangle $A C D$ is $\frac{\sqrt{3}}{2}$, let the distance from $B$ to the plane $A C D$ be $h$, from $\frac{1}{3} h \cdot S_{A C D}=\frac{1}{6}$, we get $h=\frac{\sqrt{3}}{3}$.