Example 4. Construct an inscribed parallelogram in $\triangle \mathrm{ABC}$, such that two of its sides coincide with two sides of the triangle, and one vertex lies on the third side of the triangle. At what position of point $\mathrm{P}$ will the area of the resulting parallelogram be maximized?
Analysis: As shown in Figure 4, regardless of how point $\mathrm{P}$ moves on $\mathbf{AC}$, the angle $\angle \mathrm{B}$ of the parallelogram $\mathrm{BRPQ}$ remains constant. Therefore, the area of parallelogram $\mathrm{BRPQ}$ changes with the variations of $B R$ and $B Q$.

Solution $\mathrm{S}_{a} \cdot \Delta \times \cdot \mathrm{Q}=\mathrm{BR} \cdot \mathrm{BQ} \cdot \sin \mathrm{B}$. By the parallel intercept theorem, $\mathrm{BR}: \mathrm{AP}=\mathrm{BC}: \mathrm{AC}$, and $\mathrm{BC} : \mathrm{AC}$ is a fixed ratio, so $B R$ varies with $\mathrm{AP}$, and $\mathrm{BQ}$ varies with $\mathrm{CP}$. Therefore, $\mathrm{BR} \cdot \mathrm{BQ}$ varies with $\mathrm{AP} \cdot \mathrm{CP}$. From Example 3, we know that when $\mathrm{P}$ is at the midpoint of $\mathrm{AC}$, $A P \cdot C P$ reaches its maximum value, at which point $Q$ is also at the midpoint of $A B$. Therefore, the area of $\square \mathrm{BRPQ}$ is maximized:
$$
S_{\square \# \# O}=\frac{1}{2} S_{\triangle \Delta A c} \text{. }
$$

The above is about finding the maximum value. If a constant is subtracted from a maximum value, the minimum value is obtained.