Problem 19. In triangle $\boldsymbol{A B C}$, draw a line $M N$ parallel to the base $A C$ so that the area of rectangle MNPQ is maximized.
S o l u t i o n.

Let $|A C|=a,|M N|=x,|Q M|=y,|D B|=H$ (Fig. 25). Since $\triangle A B C \sim \triangle M B N$, we have $\frac{|M N|}{|A C|}=\frac{|B O|}{|B D|}$, or $\frac{x}{a}=\frac{H-y}{H}$.

![](https://cdn.mathpix.com/cropped/2024_05_21_07b5a7f6f561b0ff634dg-23.jpg?height=337&width=437&top_left_y=2036&top_left_x=341)

Fig. 25 From this, $x=\frac{(H-y) a}{H}$. The area of rectangle $M N P Q$ is $S=x \cdot y=\frac{a}{H} \cdot(H-y) \cdot y$.

$S$ reaches its maximum value if $H-y=y$, i.e., $y=\frac{H}{2}$. Therefore, $[M N]$ is the midline of $\triangle A B C$. 22