Example 1.1.1 In $\triangle A B C$, the base $B C=a$ and the altitude $h_{a}$ to side $B C$ are both constants. When does the product of the three altitudes $h_{a} h_{b} h_{c}$ reach its maximum value?
Analyzing vertex $A$ on a line $l$ parallel to $BC$ and at a distance $h_{a}$. We might conjecture that $h_{a} h_{b} h_{c}$ is maximized when $AB=AC$; or when $\angle BAC=90^{\circ}$.

We can start from familiar area relationships to reduce the problem: In our problem, $h_{a}$ is known, so we only need to find the maximum value of $h_{b} h_{c}$; noting that the area $S=\frac{1}{2} h_{a} a$ is known, and $bc=\frac{4 S^{2}}{h_{b} h_{c}}$, the problem can be reduced to finding the minimum value of $bc$.
And $bc=\frac{2 S}{\sin A}$, which can be reduced to finding the maximum value of $\sin A$.
At this point, you can draw one or two diagrams and make the corresponding judgments.
Solution If $\angle BAC$ can be a $90^{\circ}$ angle, then when $\angle BAC=90^{\circ}$, $\sin A$ is maximized. $h_{u} h_{b} h_{c}=h_{u} \cdot \frac{4 S^{2}}{bc}=h_{u} \cdot 2 S \cdot \sin A=a h_{a}^{2} \cdot \sin A$ is maximized. At this time, the circle with $BC$ as the diameter intersects the line $l$, $h_{a} \leqslant a / 2$, and $h_{a} h_{b} h_{c}$ has a maximum value of $a h_{a}^{2}$ when $\angle BAC$ is a right angle.

If $h_{a}>a / 2$, $\angle BAC$ cannot be $90^{\circ}$, then point $A$ is outside the circle with $BC$ as the diameter, $\angle BAC<90^{\circ}$. At this time, we find that $\sin A$ increases with $\angle A$, thus, $h_{a} h_{b} h_{c}=a h_{a}^{2} \cdot \sin A$ is maximized when point $A$ is on the perpendicular bisector of $BC$, i.e., when $AB=AC$, $\angle A$ is maximized, and $\sin A$ is also maximized. That is, $h_{a} h_{b} h_{c}$ is maximized when $AB=AC$.