Test $B-5$. Let $T$ be the inscribed trapezoid $ABCD$ (counterclockwise) in the unit circle $O$, $ABCDI, AB=s_{1}$, $CD=s_{2}, OE=d, E$ is the intersection point of the heights. When $d \neq 0$, determine the minimum upper bound of $\frac{s_{1}-s_{2}}{d}$. If the minimum upper bound can be achieved, determine all such cases.
Solve: As shown in the figure, establish a rectangular coordinate system with $O$ as the origin, $AB \perp x$-axis, and the coordinates of $E$ are $(d, 0)$. The equation of $BD$ can be set as $x - d = k y$, where $k^{-1}$ is the slope of line $BD$. Since $B$ and $D$ are on the unit circle, we have
$$
(d + k y)^{2} + y^{2} = 1.
$$

Thus, $\left(k^{2} + 1\right) y^{2} + 2 d k y + \left(d^{2} - 1\right) = 0$.
As shown in the figure, let $B\left(x_{1}, y_{1}\right)$ and $D\left(x_{2}, y_{2}\right)$, then $y_{1} > y_{2}$ are the roots of (5), and $y_{1} > 0, y_{2} < 0$, so
$$
\begin{array}{l}
s_{1} - s_{2} = -2 y_{1} - 2 y_{2} = \frac{4 d k}{k^{2} + 1}, \\
\frac{s_{1} - s_{2}}{d} = \frac{4 k}{k^{2} + 1} \leqslant 2.
\end{array}
$$
The upper bound 2 in (6) is reached if and only if $k = 1$, which means that line $BD$ forms an angle of $\frac{\pi}{4}$ with the $x$-axis. By symmetry, this also means that $AC \perp BD$ (but the intersection point $E$ does not coincide with point $O$) when it is reached.