Quadrilateral BCDK and pentagon ABCDE.

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.

  • 1
    $\begingroup$ Think I cracked it. Is the answer $\sqrt3$? $\endgroup$
    – user808951
    Commented Jan 23, 2021 at 17:32
  • 1
    $\begingroup$ Yeah I got same. @Student1058 $\endgroup$
    – cosmo5
    Commented Jan 23, 2021 at 17:36
  • $\begingroup$ @Student1058 Yes, the answer is $\sqrt 3$ $\endgroup$
    – Limestone
    Commented Jan 24, 2021 at 3:53

1 Answer 1


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

  • 2
    $\begingroup$ Thank you very much for the solution. Really appreciate it. $\endgroup$
    – Limestone
    Commented Jan 24, 2021 at 3:57

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