Example 6 Let $a, b, c, d$ be positive real numbers, $a < b$ and $c < d$. There is a triangle with side lengths $\sqrt{a^{2}+c^{2}}, \sqrt{b^{2}+d^{2}}, \sqrt{(b-a)^{2}+(d-c)^{2}}$. Then the area of this triangle is $\qquad$
(12th Five Sheep Cup Junior High School Mathematics Competition)
Solution: As shown in Figure 3, construct rectangle $ABCD$ such that $AB = b - a$, $AD = c$. Extend $DA$ to point $E$ such that $DE = d$, and extend $DC$ to point $F$ such that $DF = b$. Connect $EF$ and $FB$. Then,
$$
\begin{array}{l}
BF = \sqrt{a^2 + c^2}, \quad EF = \sqrt{b^2 + d^2}, \\
BE = \sqrt{(b-a)^2 + (d-c)^2}.
\end{array}
$$

Thus, $\triangle BEF$ is the triangle given in the problem. Therefore,
$$
\begin{array}{l}
S_{\triangle BEF} = S_{\text{rectangle }ABCD} + S_{\triangle BCF} + S_{\triangle ABE} - S_{\triangle DEF} \\
= (b-a)c + \frac{1}{2}ac + \frac{1}{2}(d-c)(b-a) - \frac{1}{2}bd \\
= \frac{1}{2}(bc - ad)
\end{array}
$$