Solve:
As shown in the figure, $120^{\circ}-\theta+A+\varphi=180^{\circ} \Rightarrow \varphi=60^{\circ}-A+\theta$,
$$
\begin{array}{l}
E C=x \cos \theta, \frac{A E}{\sin \left(60^{\circ}-A+\theta\right)}=\frac{x}{\sin A} \\
\Rightarrow A E=\frac{x \sin \left(60^{\circ}-A+\theta\right)}{\sin A} \text {, thus } \\
b=x \cos \theta+\frac{x \sin \left(60^{\circ}-A+\theta\right)}{\sin A} \\
\Rightarrow x=\frac{b}{\cos \theta+\frac{\sin \left(60^{\circ}-A+\theta\right)}{\sin A}}=\frac{b \sin A}{\sin A \cos \theta+\sin \left(60^{\circ}-A+\theta\right)} \\
=\frac{b \sin A}{\left[\sin A+\sin \left(60^{\circ}-A\right)\right] \cos \theta+\cos \left(60^{\circ}-A\right) \sin \theta} \\
=\frac{b \sin A}{\left(\frac{\sqrt{3}}{2} \cos A+\frac{1}{2} \sin A\right) \cos \theta+\left(\frac{1}{2} \cos A+\frac{\sqrt{3}}{2} \sin A\right) \sin \theta} \\
\geqslant \frac{b \sin A}{\sqrt{\left(\frac{\sqrt{3}}{2} \cos A+\frac{1}{2} \sin A\right)^{2}+\left(\frac{1}{2} \cos A+\frac{\sqrt{3}}{2} \sin A\right)^{2}}}=\frac{b \sin A}{\sqrt{1+\sqrt{3} \sin A \cos A}}, \\
\end{array}
$$

Then $S_{\triangle D E F}=\frac{\sqrt{3}}{4} x^{2} \geqslant \frac{\sqrt{3}}{4} \cdot \frac{b^{2} \sin ^{2} A}{1+\sqrt{3} \sin A \cos A}=\frac{\sqrt{3}}{4} \cdot \frac{b^{2} \cdot \frac{a^{2}}{a^{2}+b^{2}}}{1+\sqrt{3} \cdot \frac{a b}{a^{2}+b^{2}}}=\frac{\sqrt{3} a^{2} b^{2}}{4\left(a^{2}+b^{2}+\sqrt{3} a b\right)}$.
Therefore, the minimum area of $\triangle D E F$ is $\frac{\sqrt{3} a^{2} b^{2}}{4\left(a^{2}+b^{2}+\sqrt{3} a b\right)}$.
11. Given Rt $\triangle A B C$, where $\angle A C B=90^{\circ}, B C=a, C A=b$. Points $D, E, F$ are on sides $B C, C A, A B$ respectively, such that $\triangle D E F$ is an equilateral triangle. Find the minimum area of $\triangle D E F$.