9 Given $\triangle A B C$, where $D$ is the midpoint of $A C$, $A B=3, B D=B C$, and the area of $\triangle A B C$ is 3, then the size of $\angle A$ is $\qquad$ .
$9 \frac{\pi}{4}$ Hint: Draw $B E \perp A C$ at point $E$, let $B E=x, A E=y$, then $x^{2}+y^{2}=9$.
$$
S_{\triangle A B C}=\frac{1}{2} \cdot x \cdot \frac{4}{3} y=\frac{2}{3} x y=3,
$$

So $x=y=\frac{3 \sqrt{2}}{2}, \angle A=\frac{\pi}{4}$.