Given $\triangle A B C, \angle A, \angle B, \angle C$ are opposite to sides $a, b, c$ respectively. In the plane of $\triangle A B C$, draw a line $l$ intersecting two sides of $\triangle A B C$. Cut $\triangle A B C$ along $l$ to form a spatial figure, and connect the smaller triangles formed by $l$ to create a two-sided pyramid. Ask:
(1) When $a=b=c$, how should $l$ be drawn, and what kind of pyramid should be formed to maximize the volume of the resulting pyramid? (Detailed proof required)
(2) When $a \leqslant b \leqslant c$, how should $l$ be drawn, and what kind of pyramid should be formed to maximize the volume of the resulting pyramid? (State the result, no proof required)
(1) Given the volume of the pyramid $=\frac{1}{3} \times$ height $\times$ base area, it is easy to know that to maximize the volume of the pyramid, the two parts of the broken line must be perpendicular.
$$
A H=\frac{2 S \Delta \Delta D B}{D E}=\frac{\sqrt{3} x y}{2 \sqrt{x^{2}+y^{2}-x y}} .
$$

Thus, when $\triangle A D E$ is folded along $D E$ and forms a right dihedral angle with $B C D E$, the volume of the formed pyramid is
$$
\begin{aligned}
V & =\frac{1}{3}\left(\frac{\sqrt{3}}{4} a^{2}-\frac{\sqrt{3}}{4} x y\right) \\
& \cdot \frac{\sqrt{3} x y}{2 \sqrt{x^{2}+y^{2}}-x y} \\
& \leqslant \frac{1}{8}\left(a^{2}-x y\right) \frac{x y}{\sqrt{2 x y-x y}} \\
& =\frac{1}{8} \sqrt{\frac{1}{2}\left(a^{2}-x y\right)^{2} 2 x y} \\
\leqslant \frac{1}{8} & \sqrt{\frac{1}{2}\left(\frac{\left(a^{2}-x y\right)+\left(a^{2}-x y\right)+2 x y}{3}\right)^{3}} \\
& =\frac{a^{3}}{12 \sqrt{3}} \\
V & =\frac{a^{8}}{12 \sqrt{3}} \text { when and only when } x=y \text { and } a^{2}-x y
\end{aligned}
$$
$=2 x y$, i.e., $x=y=\frac{a}{\sqrt{3}}$.
Therefore, when $l$ intersects $A B, A C$ at $E, D$, $A E=A D=\frac{a}{\sqrt{3}}$, and the plane of $\triangle A D E$ is perpendicular to the plane of quadrilateral $B C D E$, the volume of the pyramid is maximized.
(2) When $l$ intersects $A B, A C$ at $E, D$, $A E=A D=\frac{\sqrt{b_{c}}}{\sqrt{3}}$, and the plane of $\triangle A D E$ is perpendicular to the plane of quadrilateral $B C D E$, the volume of the pyramid is maximized.
Proof for (2):
Divide into three discussions.
First step: Prove that the $l$ that satisfies the above conditions must divide $\triangle A B C$ into an isosceles triangle with $l$ as the base.
1. As shown in the figure, if $\sqrt{\frac{2 S \triangle \triangle D B}{\sin C}} \leqslant \min\{A C$, $B C\}$, it needs to be proven that when $x=y$, $h_{1}$ is the maximum.
2. If $\sqrt{\frac{2 S \triangle \mathrm{CDB}}{\sin C}}>\min \{A C, B C\}$ $=a=B CB C$, then $C D>B C$. Therefore, when $E$ coincides with $B$, $h_{1}$ is the maximum.
3. If considered from point $A$ or point $B$, similar results can also be obtained.

Second step: Prove that when $l$ is in the required position, $l$ must intersect the two adjacent sides of $A$ at $D, E$, and $A D=A E$.
Third step: Discuss the specific position of $l$.