Example 11 (2003 University of South Carolina High School Competition Problem) As shown in Figure 5-10, extend the three sides of $\triangle ABC$ so that $BD=\frac{1}{2} AB, CE=\frac{1}{2} BC, AF=\frac{1}{2} CA$. Then the ratio of the area of $\triangle DEF$ to the area of $\triangle ABC$ is ( ).
A. $3: 1$
B. $13: 4$
C. $7: 2$
D. $14: 15$
E. $10: 3$
Solution: Choose B. Reason: As shown in the figure, draw the altitude from point $C$ to side $AB$ of $\triangle ABC$, with the foot of the perpendicular at point $X$. Draw the altitude from point $F$ to side $AD$ of $\triangle ADF$, with the foot of the perpendicular at point $Y$. Then
$\triangle AXC \sim \triangle AYF$.
Thus, $\frac{FY}{CX}=\frac{AF}{AC}=\frac{1}{2}$.
$S_{\triangle ADF}=\frac{1}{2} \times AD \times FY=\frac{1}{2} \times \frac{3}{2} \times AB \times \frac{1}{2} CX=\frac{3}{4}\left(\frac{1}{2} \times AB \times CX\right)=\frac{3}{4} S_{\triangle ABC}$.
Similarly, $S_{\triangle BED}=S_{\triangle CEF}=\frac{3}{4} S_{\triangle ABC}$.
Therefore, $S_{\triangle DEF}=S_{\triangle ABC}+S_{\triangle BED}+S_{\triangle CFE}+S_{\triangle ADF}=\left(1+\frac{3}{4}+\frac{3}{4}+\frac{3}{4}\right) S_{\triangle ABC}=\frac{13}{4} S_{\triangle ABC}$.
Thus, the required ratio is $\frac{13}{4}$.