Example 6 As shown in Figure 10, in $\triangle ABC$, $AB=AC$,
$\angle BAC=120^{\circ}$,
$\triangle ADE$ is an equilateral
triangle, point $D$ is on
side $BC$. It is known that
$BD: DC=2: 3$. When the area of $\triangle ABC$ is $50 \mathrm{~cm}^{2}$, find the area of $\triangle ADE$.
(7th Japan Arithmetic Olympiad (Final))
Analysis: Directly solving the problem is difficult. By rotating $\triangle A B C$ counterclockwise around point $A$ by $120^{\circ}$ and $240^{\circ}$ to form a regular $\triangle M B C$ (as shown in Figure 11), the regular $\triangle A D E$ becomes regular $\triangle A D_{1} E_{1}$ and regular $\triangle A D_{2} E_{2}$. It is easy to see that the hexagon $D E D_{1} E_{1} D_{2} E_{2}$ is a regular hexagon, and $\triangle D D_{1} D_{2}$ is a regular triangle, whose area is 3 times the area of $\triangle A D E$. Therefore, by finding the area of $\triangle D D_{1} D_{2}$ from the area of the regular $\triangle M B C$ which is 150, the problem is solved.

Solution: Note that $B D: D C=C D_{1}: D_{1} M=M D_{2}: D_{2} B=2: 3$. Connecting $D M$, we have
$S_{\triangle M B D}=\frac{2}{5} S_{\triangle M B C}=60\left(\mathrm{~cm}^{2}\right)$.
And $S_{\triangle D_{2} B D}=\frac{3}{5} S_{\triangle M B D}=36\left(\mathrm{~cm}^{2}\right)$.
Similarly, $S_{\triangle W D_{1} D_{2}}=S_{\triangle \Delta D_{1}}=36\left(\mathrm{~cm}^{2}\right)$.
Therefore, $S_{\triangle D D_{1} D_{2}}=150-3 \times 36=42\left(\mathrm{~cm}^{2}\right)$.
Thus, $S_{\triangle A D E}=\frac{1}{3} S_{\triangle D D_{1} D_{2}}=14\left(\mathrm{~cm}^{2}\right)$.