14. (9) In parallelogram $A B C D$, angle $\angle B A C$ is twice the angle $\angle B D C$. Find the area of the parallelogram, given that $A B=$ $A C=2$.
Answer: $2 \sqrt{3}$.

Solution: The specified parallelogram is a rhombus with an angle of 60 degrees. This can be proven by constructing a circle centered at point $A$ with a radius of 2. Then points $B, C, D$ will lie on this circle. The side of the rhombus is 2, so its area $S(A B C D)=A B \cdot A D$.

$\sin 60^{\circ}=2 \sqrt{3}$

![](https://cdn.mathpix.com/cropped/2024_05_06_7a0901b1302859fc60d1g-05.jpg?height=475&width=442&top_left_y=1367&top_left_x=798)

Since a typo was introduced in the published condition of this problem - $\angle B D C$ was specified instead of $\angle B C D$, a correct proof that the construction described in the problem does not exist is also accepted as a correct solution.