24. In an isosceles $\triangle ABC$ with base angles $\alpha$ and leg length $a$, every inscribed triangle in $\triangle ABC$ has a largest side (in this problem, an inscribed triangle refers to one where each side of $\triangle ABC$ contains at least one vertex of the inscribed triangle). Find the minimum value of these largest sides, and determine the shape of this inscribed triangle at that time.
$24.1^{\circ}$ From Figure 13-20(1), it is not difficult to deduce that for all $\triangle D E F$ with a given side $E F$, the triangle with the minimum value of the maximum side is the isosceles $\triangle D^{\prime} E F$. Therefore, we only need to discuss the isosceles $\triangle D E F (D E = D F, D$ on $B C$ side).
$2^{\circ}$ Take $O$ as the midpoint of $E F$ (Figure 13-20(1), (2)), draw $G H / / B C$, and draw $G P / / A E$, then $H E = G F = P G$. Thus, $\angle F P E = \angle F Q H = \angle G P F + \angle G H A > \angle G P F + \angle F E H = \angle P F E$. Therefore, $G H \sqrt{O G^{2} + O D^{2}} = D G$.
So, the original problem is further reduced to discussing isosceles triangles with the base parallel to $B C$ and the vertex on $B C$ side (i.e., the midpoint).
$3^{\circ}$ Let $\angle D A C = \alpha, \angle F D C = \beta, D F = x$, $A D = h, B C = a$. If $\alpha \geqslant 60^{\circ}$, draw $D F \perp A C, D E \perp A B, E, F$ as the feet of the perpendiculars, then $D F = \frac{a}{2} \cos \alpha$ is the maximum side of $\triangle D E F$ ( $D F > 2 D F \cos \alpha = E F$). Since the value increases when $F$ is in other positions, $\frac{a}{2} \cos \alpha$ is the minimum value sought.

If $\alpha < 60^{\circ}$, $x$ is still the maximum side (as shown in Figure 13-20(5)). Moving $F$ towards $C$ (keeping $F E$ parallel to $B C$), $\beta, x$ decrease accordingly, until $\beta = 60^{\circ}$, at which point $x = E F$. Therefore, we only need to consider the case where $\beta \leqslant 60^{\circ}$. At this time,
$$
\frac{2 x \cos \beta}{a} = \frac{h - x \sin \beta}{h}, \text{ i.e., } \frac{2 x \cos \beta}{a} = \frac{2 h}{a \operatorname{tg} \beta + 2 h}.
$$

The maximum side $E F = 2 x \cos \beta = \frac{2 a h}{a \operatorname{tg} \beta + 2 h}$ decreases as $\beta$ increases, and reaches its minimum value $\frac{2 a h}{a \sqrt{3} + 2 h}$ when $\beta = 60^{\circ}$, at which point $\triangle D E F$ is an equilateral triangle.