Example 3. Take a point $P$ inside $\triangle ABC$, and draw three lines through $P$ parallel to the three sides of $\triangle ABC$,
thus forming three triangles
$t_{1}, t_{2}, t_{3}$ with areas
4, 9, 49 (as
shown in Figure 3).
Find the area of $\triangle ABC$.

(From the 2nd $\triangle I M E$)
$$
\begin{array}{l}
\text{Given that } \triangle t_{1}, \triangle t_{2}, \triangle t_{3} \text{ and } \triangle A B C \text{ are all similar triangles.} \\
\text{Let the areas of } \triangle A B C, \triangle t_{1}, \triangle t_{2}, \triangle t_{3} \text{ be } S, S_{1}, S_{2}, S_{3} \text{, respectively.} \\
\text{It is easy to know,} \\
\frac{D A}{C A}=\frac{P I}{C A}=\frac{\sqrt{S_{1}}}{\sqrt{S}}, \frac{C G}{C A}=\frac{H P}{C A} \\
=\frac{\sqrt{S_{2}}}{\sqrt{S}}, \quad G D=\sqrt{S_{3}} . \\
\because D A+C G+G D=C A, \\
\therefore \quad \frac{\sqrt{S_{1}}}{\sqrt{S}}+\frac{\sqrt{S_{2}}}{\sqrt{S}}+\frac{\sqrt{S_{3}}}{\sqrt{S}}=1. \\
\text{Thus,} \quad \sqrt{S}=\sqrt{S_{1}}+\sqrt{S_{2}}+\sqrt{S_{3}} \\
=2+3+7=12. \\
\end{array}
$$

Therefore, $S=144$.