Example 17 If $a \geqslant b>c$ and $a<b+c$, solve the equation
$$
b \sqrt{x^{2}-c^{2}}+c \sqrt{x^{2}-b^{2}}=a x \text {. }
$$
(1993, Nanjing City Junior High School Mathematics Competition)
Solution: According to the problem, we can construct $\triangle ABC$ with $a, b$, and $c$ as sides, such that $BC = a, CA = b, AB = c$, and $a \geqslant b > c$. Draw perpendiculars to $AC$ and $BA$, intersecting at point $D$. Thus, $ABDC$ is inscribed in a circle with diameter $AD$. By Ptolemy's theorem, we have
$$
AC \cdot BD + AB \cdot CD = AD \cdot BC.
$$

That is,
$$
\begin{array}{l}
b \sqrt{AD^2 - c^2} + c \sqrt{AD^2 - b^2} \\
= a AD.
\end{array}
$$

Comparing this equation with the original equation, we know
$$
AD = x.
$$

In $\triangle ABC$,
$$
\begin{array}{l}
AD = 2R = \frac{a}{\sin \angle CAB}, \\
S_{\triangle ABC} = \frac{1}{2} bc \sin \angle CAB.
\end{array}
$$

And $S_{\triangle ABC} = \sqrt{s(s-a)(s-b)(s-c)}$
$$
\begin{array}{l}
\left(s = \frac{a+b+c}{2}\right), \\
\therefore \quad AD = \frac{abc}{2 \sqrt{s(s-a)(s-b)(s-c)}},
\end{array}
$$

Thus, $x = \frac{abc}{2 \sqrt{s(s-a)(s-b)(s-c)}}$.