Example 7 (5th "Hope Cup" Invitational Competition Question) $D$ and $E$ are the trisection points of the edges $AB$ and $AC$ of a regular tetrahedron (a triangular pyramid with all edges of equal length) $V-ABC$ ($AD: DB=2: 1, AE: EC=2: 1$), then the section through $D$ and $E$ and parallel to $VA$ is ( ).
A. Square
B. Rectangle but not a rhombus
C. Rhombus but not a rectangle
D. Parallelogram but not a rectangle or a rhombus
Solution: Choose B. Reason: As shown in Figure 7-7, in $\triangle ABC$, $AD : DB = AE : EC = 2 : 1$, which implies $DE \parallel BC$, and $DE = \frac{2}{3} BC$. Given that $VA \parallel$ the section $DEFG$, it follows that $DG \parallel EF, DG = EF$, and $DG = \frac{1}{3} VA$, making $DEFG$ a parallelogram with unequal adjacent sides. Since $VA = VB = VC = AB = BC = CA$, it follows that $VA \perp DE$, hence $DG \perp DE$, and $DEFG$ is a rectangle but not a rhombus.