# Find the area of this pentagon

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.

• Think I cracked it. Is the answer $\sqrt3$?
– user808951
Commented Jan 23, 2021 at 17:32
• Yeah I got same. @Student1058 Commented Jan 23, 2021 at 17:36
• @Student1058 Yes, the answer is $\sqrt 3$ Commented Jan 24, 2021 at 3:53

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 $$\color{blue}{\sqrt3}$$.