Four. (25 points) As shown in Figure 2, in the right trapezoid $ABCD$, $\angle ABC = \angle BAD = 90^{\circ}$, $AB = 16$. The diagonals $AC$ and $BD$ intersect at point $E$. A line $EF \perp AB$ is drawn through $E$ at point $F$, and $O$ is the midpoint of side $AB$, with $FE + EO = 8$. Find the value of $AD + BC$.
Let $O F=x$.
Then $F B=8-x, F A=8+x$.
Given that $D A / / E F / / C B$, we have $\frac{F E}{A D}=\frac{F B}{A B}$, which means
$$
A D=\frac{16}{8-x} E F \text {. }
$$

Similarly, $B C=\frac{16}{8+x} E F$.
Thus, $A D+B C=\left(\frac{16}{8-x}+\frac{16}{8+x}\right) E F$
$$
=\frac{16 \times 16}{8^{2}-x^{2}} E F \text {. }
$$

Also, in $\mathrm{RI} \triangle E F O$ †, $E F^{2}+O F^{2}=O E^{2}$, which means $E F^{2}+x^{2}=(8-E F)^{2}$.
Solving this, we get $E F=\frac{8^{2}-x^{2}}{16}$.
From equation (D).(2), we have $A D+B C=16$.