Prove that if $0 \leq a < b$ and $0 \leq c < d$, then $ac < bd$.

Question

This is an exercise from Spivak's "Calculus". Is it possible to do this problem without using many cases?

If $0 \leq a < b$ and $0 \leq c < d$, then $ac < bd$.

My solution references the result of a previous problem:

5.iv If $a < b$ and $c > 0$, then $ac < bc$.

Proof: Suppose $0 \leq a < b$ and $0 \leq c < d$. We then consider cases.
Case 1. $a = 0$. We then consider two subcases since $c \geq 0$.
     Case 1.a. $c = 0$. Then since $b > 0$ and $d > 0$, $bd > 0$. Thus $bd > ac$.
     Case 1.b. $c > 0$. Similary, $ac = (0)c = 0 < bd$. So $ac < bd$
Case 2. $a > 0$. We again consider two subcases.
     Case 2.a. $c = 0$. $ac = a(0) = 0 < bd$.
     Case 2.b. $c > 0$. Then $ac > 0$. By 5.iv, $ac < bc$. Again by 5.iv, $bc < bd$. Combining these two           inequalities gives us $ac < bc < bd$. Thus, $ac < bd$.
In every case, we have shown that $ac < bd$. $\square$

Answer 1

If $a=0$ or $c=0$, then $ac=0<bd$.

Otherwise, $ac<ad<bd$.

Answer 2

$bd-ac=bd-ad+ad-ac=d(b-a)+a(d-c)>0.$

The last step follows from the hypothesis.