Problem 1385

1. Given that $AB$ extended and $CD$ extended intersect at a right angle, we can set the y-axis as $\overline{AB}$ and the x-axis as $\overline{CD}$. This means that $A$ is at $(0, a)$, $B$ is at $(0, b)$, $C$ is at $(c, 0)$, and $D$ is at $(d, 0)$.

2. We need to prove that $AC \cdot BD > AD \cdot BC$. First, we calculate the lengths of the diagonals $AC$, $BD$, $AD$, and $BC$:
$$AC = \sqrt{a^2 + c^2}, \quad BD = \sqrt{b^2 + d^2}, \quad AD = \sqrt{a^2 + d^2}, \quad BC = \sqrt{b^2 + c^2}$$

3. We then compute the products $AC \cdot BD$ and $AD \cdot BC$:
$$AC \cdot BD = \sqrt{(a^2 + c^2)(b^2 + d^2)}$$
$$AD \cdot BC = \sqrt{(a^2 + d^2)(b^2 + c^2)}$$

4. To prove $AC \cdot BD > AD \cdot BC$, we need to show:
$$(a^2 + c^2)(b^2 + d^2) > (a^2 + d^2)(b^2 + c^2)$$

5. Expanding both sides, we get:
$$a^2b^2 + a^2d^2 + b^2c^2 + c^2d^2 > a^2b^2 + a^2c^2 + b^2d^2 + c^2d^2$$

6. Simplifying, we need to show:
$$a^2d^2 + b^2c^2 > a^2c^2 + b^2d^2$$

7. Rearranging terms, we get:
$$a^2(d^2 - c^2) > b^2(d^2 - c^2)$$

8. We consider two cases:
- If $a < b$ and $d < c$, then $d^2 - c^2 < 0$. Thus, it suffices to show that $a^2 < b^2$, which holds since $a < b$.
- If $a > b$ and $d > c$, then $d^2 - c^2 > 0$. Thus, it suffices to show that $a^2 > b^2$, which holds since $a > b$.

Therefore, in both cases, the inequality $a^2(d^2 - c^2) > b^2(d^2 - c^2)$ holds, proving that $AC \cdot BD > AD \cdot BC$.

$\blacksquare$