A Simple Geometrical Inequality

Let $O$ be the intersection point of $AC$ and $BD$ such that $OC = x, OD = y$ . Since $AB \parallel CD$ , we have similar right triangles $AOB, COD$ such that:

$\frac{OC}{OD} = \frac{OA}{OB} \Rightarrow \frac{x}{y} = \frac{40-x}{30-y} \Rightarrow y = \frac{3}{4}x.$

To determine the minimum length of $AD$ , we use the Pythagorean Theorem on right triangle $AOD$ :

$|AD| = \sqrt{(OA)^2 + (OD)^2} = \sqrt{(40-x)^2 + (\frac{3}{4}x)^2} = \sqrt{1600 - 80x + \frac{25}{16}x^2} = \sqrt{\frac{25}{16}(x - \frac{128}{5})^2 + 576}.$

Since the radicand is a concave-up parabola with minimum value of 576, the minimum length of $|AD|$ is $\sqrt{576} = \boxed{24}.$