Example 3.8.4 (CMO 1996) In $\triangle A B C$, $\angle C=90^{\circ}, \angle A=30^{\circ}, B C=1$, find the minimum value of the longest side of the inscribed triangle in $\triangle A B C$.
(1) On $BC$, take $BD=x$, and take points $E$ and $F$ on $CA$ and $CB$ respectively, such that $CE=\frac{\sqrt{3}}{2} x, BF=1-x / 2$, as shown in the figure. Then, by the cosine rule,
$$
\begin{array}{c}
D F^{2}=x^{2}+(1-x / 2)^{2}-2 x(1-x / 2) \cos 60^{\circ}=\frac{7}{4} x^{2}-2 x+1, \\
D E^{2}=(1-x)^{2}+\left(\frac{\sqrt{3}}{2} x\right)^{2}=\frac{7}{4} x^{2}-2 x+1, \\
E F^{2}=(1+x / 2)^{2}+\left(\sqrt{3}-\frac{\sqrt{3}}{2} x\right)^{2}-2\left(\sqrt{3}-\frac{\sqrt{3}}{2} x\right) . \\
(1+x / 2) \cos 30^{\circ}=\frac{7}{4} x^{2}-2 x+1 .
\end{array}
$$

Thus, $\triangle D E F$ is an equilateral triangle. [This construction is extraordinary!]. This indicates that for any point $D$ on $BC$, an inscribed equilateral triangle can be constructed.

Let the midpoints of $CA$ and $AB$ be $N$ and $M$ respectively, and the midpoint of $BM$ be $S$. Then, as point $D$ moves from $B$ to $C$, point $E$ moves from $C$ to $N$, and point $F$ moves from $M$ to $S$.
Let the side length of $\triangle D E F$ be $a$, then $a^{2}=\frac{7}{4}\left(x^{2}-\frac{8}{7} x+\frac{4}{7}\right)=\frac{7}{4}\left(x-\frac{4}{7}\right)^{2}+\frac{3}{7}$.
It is clear that when $x=4 / 7$, the side length reaches its minimum value $\sqrt{\frac{3}{7}}$.
(2) Below, we prove that the longest side of any inscribed triangle is not less than $\sqrt{\frac{3}{7}}$.
For convenience, we introduce a Cartesian coordinate system. Let $C$ be the origin, and $CA$ be the positive $x$-axis.
Let $\triangle X Y Z$ be any inscribed triangle of $\triangle A B C$, as shown in the figure. On $BM$, take points $R_{1}$ and $R_{2}$ such that $M R_{1}=\frac{1}{3} M B, B R_{2}=\frac{3}{14} M B$, thus, $A R_{1}=\frac{2}{3} A B, B R_{2}=\frac{11}{14} B M$,
hence $y_{R_{1}}=\frac{2}{3}>\sqrt{\frac{3}{7}} ; x_{R_{2}}=\frac{11}{14} \times \frac{\sqrt{3}}{2}>\sqrt{\frac{3}{7}}$.
(1) If point $Z$ is on the segment $B R_{1}$, then $Y Z \geqslant y_{R_{1}}>\sqrt{\frac{3}{7}}$,
if point $Z$ is on the segment $R_{2} A$, then $X Z \geqslant x_{R_{2}}>\sqrt{\frac{3}{7}}$,
thus, in this case, the longest side of $\triangle X Y Z$ is $>\sqrt{\frac{3}{7}}$.
(2) If point $Z$ is inside the segment $R_{1} R_{2}$, then take point $Z$ as point $F$, and construct the equilateral $\triangle D E F$. It can be verified that: $x_{E}<x_{F}$, $y_{D}<y_{F}$. Therefore,
if point $X$ is on the segment $D C$, then $Z X \geqslant Z D=F D$;
if point $Y$ is on the segment $C E$, then $Z Y \geqslant Z E=F E$;
if point $X$ is on the segment $B D$, and point $Y$ is on the segment $E A$, then by the Pythagorean theorem
$X Y \geqslant D E . \therefore$ the longest side of $\triangle X Y Z$ is not less than the side length of $\triangle D E F$, and thus not less than $\sqrt{\frac{3}{7}}$.
In summary, the minimum value of the longest side is $\sqrt{\frac{3}{7}}$.