Question:
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 such that $AB=BC$ and $DE=DC$ and $K$ lies in the interior of the pentagon $ABCDE$. If $\angle ABC = 120 ^{\circ}$, $\angle CDE = 60^{\circ}$ and $BD=2$ then determine $\text {Area{ABCDE}}$
I did some angle chasing and concluded that $CD=CE=DE$ and that $\Delta AKE$ is right angled at $K$. But I'm not getting any idea to find the area of the Pentagon.
There are infinitely many configurations satisfying the given conditions, but in all of them the pentagon $P$ in question has the same area.
Let $|BC|=b$, $|DC|=d$, and $\angle(BCD)=\gamma$. Then $|AC|=b\sqrt{3}$, $|CE|=d$, and $\angle (ACE)=\gamma-90^\circ$. From these data one computes $$\eqalign{{\rm area}(P)&={1\over2}b^2\sin 120^\circ+{1\over2}d^2\sin60^\circ+{1\over2}\ b\sqrt{3}\ d\sin(\gamma-90^\circ)\cr &={\sqrt{3}\over4}\bigl(b^2+d^2-2bd\cos\gamma\bigr)\ .\cr}$$ Now the cosine theorem gives $$b^2+d^2-2bd\cos\gamma=|BD|^2=4\ ,$$ so that we obtain ${\rm area}(P)=\sqrt{3}$, whatever the values of $b$ and $d$ under the given circumstances.