Find the area of this pentagon

Question

Let $BCDK$ be a convex quadrilateral with $BC=BK$ and $DC=DK$. $A$ and $E$ are points such that $AB=BC$, $DE=DC$ and such that $ABCDE$ is a convex pentagon. Point $K$ lies in the interior of pentagon $ABCDE$. If $\angle ABC=120^{\circ}$, $\angle CDE=60^{\circ}$ and $BD=2$, find the area of pentagon $ABCDE$.

I know that this can be solved using trigonometry.But i prefer to have a solution by euclidean geometry over that. I tried extending some lines and finding other trivia, but they didn't seem to work.

Answer 1

Rotate $\triangle BCD$ $60$ degrees w.r.t point $D$ to $\triangle B'ED$. The brown area will be the desired area.

Connect $\overline{BB'}$. Since $\angle BDB'=60^{\circ}$ and $\overline{BD}=\overline{B'D}$, we can say that $\triangle BDB'$ is equilateral.

Assume $\overline{AE}$ and $\overline{BB'}$ intersect at $F$.

Let $\angle ABF=a$ and $\angle EB'F=b$. By symmetry, $\angle FBK=b$ and $\angle KBD=\angle CBD=60^{\circ}-b$. Therefore we have $$a+b+(60^{\circ}-b)+(60^{\circ}-b)=120^{\circ}\implies a=b$$ So $$\triangle ABF\cong\triangle EB'F\quad(A.A.S.)$$ Therefore the answer will be the area of equilateral $\triangle BDB'$, which equals $\color{blue}{\sqrt3}$.