Let $A$, $B$, $C$, $D$ be four points on a line in this order.  Suppose that $AC = 25$, $BD = 40$, and $AD = 57$. Compute $AB \cdot CD + AD \cdot BC$.

[i]Proposed by Evan Chen[/i]
1. Given the points \(A\), \(B\), \(C\), and \(D\) on a line in this order, we know the following distances:
   \[
   AC = 25, \quad BD = 40, \quad AD = 57
   \]

2. First, we calculate the distance \(CD\):
   \[
   CD = AD - AC = 57 - 25 = 32
   \]

3. Next, we calculate the distance \(BC\):
   \[
   BC = BD - CD = 40 - 32 = 8
   \]

4. Then, we calculate the distance \(AB\):
   \[
   AB = AC - BC = 25 - 8 = 17
   \]

5. Now, we need to compute \(AB \cdot CD + AD \cdot BC\):
   \[
   AB \cdot CD + AD \cdot BC = 17 \cdot 32 + 57 \cdot 8
   \]

6. We can simplify the expression step-by-step:
   \[
   17 \cdot 32 = 17 \cdot (30 + 2) = 17 \cdot 30 + 17 \cdot 2 = 510 + 34 = 544
   \]
   \[
   57 \cdot 8 = 57 \cdot (10 - 2) = 57 \cdot 10 - 57 \cdot 2 = 570 - 114 = 456
   \]

7. Adding these results together:
   \[
   544 + 456 = 1000
   \]

Thus, the value of \(AB \cdot CD + AD \cdot BC\) is \(\boxed{1000}\).