Example 4 In a plane, given six line segments $l_{1}, l_{2}, \cdots$, $l_{6}$, they are equal to the edges $A B, A C$, $A D, B C, B D, C D$ of the tetrahedron $A B C D$. Question: How to use a ruler and compass to construct a line segment equal to the altitude of the tetrahedron passing through vertex $A$?
Solution: As shown in Figure 7, $AO$ is an altitude of the tetrahedron. In the base plane $BCD$, draw $OE \perp BC$ at $E$, and $OF \perp CD$ at $F$. By the theorem of three perpendiculars, we get
$AE \perp BC, AF \perp CD$.
Now, unfold the lateral faces $ABC$ and $ACD$ onto the plane $BCD$ (as shown in Figure 8), then the lines $A_1EO$ and $A_2FO$ are perpendicular to $BC$ and $CD$, respectively. Thus, the construction method is as follows:
(1) Construct $\triangle BCD$ with sides $l_4, l_5, l_6$, and construct $\triangle A_1BC$ with sides $l_1, l_2, l_4$, and construct $\triangle A_2CD$ with sides $l_2, l_3, l_6$.
(2) Draw $A_1E \perp BC$ and $A_2F \perp CD$. Let the intersection of $A_1E$ and $A_2F$ be point $O$.
(3) Connect $CO$, draw $OX \perp CO$, and with $C$ as the center and $l_2$ as the radius, draw an arc intersecting $OX$ at $A$.
Therefore, $OA$ is the altitude of the tetrahedron.