Example 19 If $a \geqslant b \geqslant c>0$, and $a<b+c$, solve the equation $b \sqrt{x^{2}-c^{2}}+c \sqrt{x^{2}-b^{2}}=a x$.


Given $a \geqslant b \geqslant c>0$, and $a<b+c$, so $a, b, c$ can form a triangle. Construct $\triangle A B C$ such that $B C=a, A C=b, A B=c$. Draw perpendiculars from $A$ to $A C$ and $A B$, and let them intersect at point $D$. Then the quadrilateral $A B D C$ is cyclic, as shown in Figure 3-23. At this point, $A D$ is the diameter, $\sin \angle B A D=\frac{B D}{A D}=\frac{\sqrt{A D^{2}-c^{2}}}{A D}, \sin \angle C A D=\frac{C D}{A D}=\frac{\sqrt{A D^{2}-b^{2}}}{A D}, \sin \angle C A B=\frac{a}{A D}$.

Applying Ptolemy's theorem or the three-chord theorem to $A D, A C, A B$, we have $A C \cdot \sin \angle B A D + A B \cdot \sin \angle C A D = A D \cdot \sin \angle C A B$, which is $b \cdot \frac{\sqrt{A D^{2}-c^{2}}}{A D} + c \cdot \frac{\sqrt{A D^{2}-b^{2}}}{A D} = A D \cdot \frac{a}{A D}$, or $b \cdot \sqrt{A D^{2}-c^{2}} + c \cdot \sqrt{A D^{2}-b^{2}} = a \cdot A D$.

Since $S_{\triangle A B C} = \frac{1}{2} b c \cdot \sin \angle C A B = \frac{a b c}{2 A D}$, and $S_{\triangle A B C} = \sqrt{p(p-a)(p-b)(p-c)}$, where $P = \frac{1}{2}(a+b+c)$, it follows that $A D = \frac{a b c}{2 \sqrt{p(p-a)(p-b)(p-c)}}$ is the solution to the original equation.