1. Let $ABCD$ be a trapezoid with $AD \parallel BC$ and let the diagonals $AC$ and $BD$ intersect at point $P$. We need to prove that the product of the legs is at least as much as the product of the bases, i.e.,
$$AB \cdot CD \ge AD \cdot BC.$$

2. Given that the diagonals are perpendicular, we have $AC \perp BD$. Let $PA = a$, $PB = b$, $PC = c$, and $PD = d$. We need to show:
$$(a^2 + b^2)(c^2 + d^2) \ge (a^2 + d^2)(b^2 + c^2).$$

3. Expanding both sides, we get:
$$(a^2 + b^2)(c^2 + d^2) = a^2c^2 + a^2d^2 + b^2c^2 + b^2d^2,$$
and
$$(a^2 + d^2)(b^2 + c^2) = a^2b^2 + a^2c^2 + d^2b^2 + d^2c^2.$$

4. Subtracting the right-hand side from the left-hand side, we obtain:
$$(a^2c^2 + a^2d^2 + b^2c^2 + b^2d^2) - (a^2b^2 + a^2c^2 + d^2b^2 + d^2c^2) = a^2d^2 + b^2c^2 - a^2b^2 - d^2c^2.$$

5. Rearranging terms, we get:
$$a^2d^2 + b^2c^2 - a^2b^2 - d^2c^2 = (a^2 - c^2)(d^2 - b^2).$$

6. Since $AD \parallel BC$, we have the ratios $\frac{PA}{PD} = \frac{PC}{PB} = t$. This implies:
$$a = td \quad \text{and} \quad c = tb.$$

7. Substituting these into the inequality, we get:
$$(t^2d^2 - t^2b^2)(d^2 - b^2) = t^2(d^2 - b^2)^2.$$

8. Since $(d^2 - b^2)^2 \ge 0$ and $t^2 \ge 0$, it follows that:
$$t^2(d^2 - b^2)^2 \ge 0.$$

9. Therefore, the inequality holds:
$$(PA^2 + PB^2)(PC^2 + PD^2) \ge (PA^2 + PD^2)(PB^2 + PC^2).$$

$\blacksquare$