Let $A B C D$ be a right trapezoid with bases $A B$ and $C D$, and right angles at $A$ and $D$. Given that the shorter diagonal $B D$ is perpendicular to the side $B C$, determine the smallest possible value for the ratio $C D / A D$.
Let $A \widehat{B} D = B \widehat{D} C = \alpha$, as shown in the given figure. Then we have $C D = \frac{B D}{\cos \alpha}$ and $A D = B D \sin \alpha$, hence

$$
\begin{aligned}
\frac{C D}{A D} & = \frac{\frac{B D}{\cos \alpha}}{B D \sin \alpha} = \frac{1}{\sin \alpha \cos \alpha} \\
& = \frac{2}{\sin 2 \alpha} \geq 2
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_01_30c9a294a58a6e4b8190g-123.jpg?height=291&width=522&top_left_y=1605&top_left_x=1098)

Thus, the smallest value of the ratio $C D / A D$ is 2, which occurs when $\sin 2 \alpha = 1$, i.e., when $\alpha = 45^{\circ}$.