2. Given the ellipse $\frac{x^{2}}{2}+y^{2}=1$ with two foci $F_{1}$ and $F_{2}$, a chord $A B$ is drawn through the right focus $F_{2}$ with an inclination angle of $\frac{\pi}{4}$. Then the area of $\triangle A B F_{1}$ is ( ).
(A) $\frac{2 \sqrt{2}}{3}$
(B) $\frac{4}{3}$
(C) $\frac{4 \sqrt{2}}{3}$
(D) $\frac{4 \sqrt{3}}{3}$
$$
\begin{array}{c}
\because F_{2}(1,0), \\
B(0,-1), \\
F_{1}(-1,0), \\
\therefore \angle A B F_{1}=90^{\circ} . \\
\text { Also, }\left|F_{1} A\right|+ \\
\left|F_{2} A\right|=2 a=2 \sqrt{2}, \\
\therefore\left|F_{1} A\right|=2 \sqrt{2}-\left|F_{2} A\right| . \\
\text { Also, }\left|F_{1} A\right|^{2}=\left|F_{1} B\right|^{2}+|A B|^{2}, \\
\therefore\left(2 \sqrt{2}-\left|F_{2} A\right|\right)^{2} \\
=\left|F_{1} B\right|^{2}+\left(\sqrt{2}+\left|F_{2} A\right|\right)^{2} . \\
\therefore\left|F_{2} A\right|=\frac{\sqrt{2}}{3},|A B|=\frac{4 \sqrt{2}}{3} . \\
\text { Then } S_{\triangle A A F_{1}}=\frac{1}{2}|A B| \cdot\left|B F_{1}\right|=\frac{4}{3} .
\end{array}
$$