Example 11. Isosceles $\triangle A O B$, vertex angle $2 \theta$, altitude on the base $h$, and the altitude is in the positive direction of the $x$-axis. Inside $\triangle A O B$ there is a point $\mathrm{P}$, its distances to $\mathrm{OA}, \mathrm{AB}, \mathrm{OB}$ are $|\mathrm{PD}|,|\mathrm{PE}|,|\mathrm{PF}|$ respectively. If $|\mathrm{PD}| \cdot|\mathrm{PF}| = |P E|^{2}$, find the equation of the locus of point $P$.
As shown in Figure 11, take the positive half-axis of $\mathrm{O} \times$ as the polar axis, and let the coordinates of the moving point $\mathrm{P}$ be $(\rho, \alpha)$.
$\because \mathrm{P}$ is inside $\triangle \mathrm{AOB}$,

\[
\begin{array}{l}
\therefore |\alpha| < \theta, \rho > 0, \\
|P D| = \rho \cdot \sin (\theta - \alpha), \\
|P F| = \rho \cdot \sin (\theta + \alpha), \\
|P E| = |h - \rho \cdot \cos \alpha|.
\end{array}
\]

By the given condition, $\rho \sin (\theta - \alpha) \cdot \rho \sin (\theta + \alpha)$
\[
= (h - \rho \cos \alpha)^{2} \quad (\text{where } |\alpha| < \theta),
\]
\[
\begin{array}{l}
\frac{1}{2} \rho^{2} (\cos 2 \alpha - \cos 2 \theta) \\
= h^{2} - 2 h \rho \cos \alpha + \rho^{2} \cos^{2} \alpha,
\end{array}
\]

Thus, $\rho^{2} \cos^{2} \theta - 2 h \rho \cos \alpha + h^{2} = 0$. (where $\theta$ is a fixed angle, $|\alpha| < \theta$)
Converting to Cartesian coordinates, the equation becomes:
\[
\cos^{2} \theta \left(x^{2} + y^{2}\right) - 2 h x + h^{2} = 0.
\]
(The segment of a circular arc inside $\triangle \mathrm{AOB}$)