Let $BCDK$ be a convex quadrilateral such that $BC$ = $BK$ and $DC$ = $DK$.
$A$ and $E$ are points such that $ABCDE$ is a convex pentagon and $AB$ = $BC$ and $DE$ = $DC$ and $K$ lies in the interior of the pentagon.
Angle $ABC$ = 120 and angle $CDE$ = 60 and $BD$ = 2 , then determine area of the pentagon $ABCDE$.
Note that $BCDK$ is a parallelogram. Because $ABCDE$ is convex, the area of $ABCDE$, shown by $S(ABCD)$, is equal to \begin{equation} S(ABCD) = S(ABC) + S(CDE) + S(ACE) \end{equation} Let's assume $CD = a$ and $BC = b$. Then, $S(ABC) = \frac{b^2\sin120}{2}$ as $AB = BC = b$ and $\hat{ABC}=120$. Similarly, we have $S(CDE) = \frac{a^2\sin60}{2}$. We use the same trick to find area of $ACE$. We get $S(ACE) = \frac{AC.CE\sin(\hat{ACE}-(30+60))}{2}$, in which 30 and 60 are angles $\hat{BCA}$ and $\hat{ECD}$ respectively (they can be found considering the fact that triangles $ABC$ and $CED$ are isosceles). Because $ABC$ and $CED$ are isosceles, we can easily show that $AC = 2b\sin60$ and $CE = a$ (in fact CED is equilateral). So, put it together we get: \begin{equation}\begin{split} S(ABCD) &= S(ABC) + S(CDE) + S(ACE)\\ &+ \frac{b^2\sin120 + a^2\sin60 + 2ab\sin60\sin(\hat{ACE}-90)}{2} \end{split} \end{equation} Considering the fact that $\sin120 = \sin 60$ and $sin(\hat{ACE}-90) =-\cos(\hat{ACE})$, we get \begin{equation} S(ABCD) = \frac{b^2 + a^2 - 2ab\cos(\hat{ACE})}{2}\sin60 \end{equation} Using the law of cosines, numerator is $BD^2$ which is 4. Thus, $S(ABCD) = 2\sin(60) = \sqrt3$.