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 $\frac{C D}{A D}$.
Let $A \widehat{B} D=B \widehat{D} C=\alpha$. Then we have that $D C=\frac{B D}{\cos \alpha}$ and $A D=B D \sin \alpha$, hence

$\frac{D C}{A D}=\frac{\frac{B D}{\cos \alpha}}{B D \sin \alpha}=\frac{1}{\sin \alpha \cos \alpha}=\frac{2}{\sin 2 \alpha} \geq 2$.

Equality occurs when $\sin 2 \alpha=1$, that is, when $\alpha=45^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_01_a444074f1f8ae50444d5g-42.jpg?height=326&width=576&top_left_y=1935&top_left_x=1143)