Fix points $ D,A,B$, rays $ BC$ and $ AE$. Try to chose points $ C$ and $ E$ on these rays so that $ \angle EDC=60^{\circ}$ and area of $ ABCDE$ takes maximal possible value. A point $ C$ varies on a segment $ BC_0$, where $ \angle ADC_0=60^{\circ}$, point $ E$ depends on $ C$ continiously. So, the minimum is attained by Weierstrass theorem. If it appears in position $ C$, $ E$, where $ C\ne B$, $ E\ne A$, then we must have $ DE=DC$: else if, say, $ DE<DC$ chose $ E'$ on segment $ AE$ close to $ E$ and $ C'$ on the ray $ CC_0$ such that $ \angle EDE'=\angle CDC'=\epsilon$. Then area($ ABCDE$)-area($ ABC'DE'$)=area($ EDE'$)-area($ CDC')=\frac{\sin\epsilon}2(DE\cdot DE'-DC\cdot DC')<0$, a contradiction with maximality. Other possibility is that this maximum is attained when, say $ E=A$. Consider the second case. We have $ ABCDE=ADCB$ is a cyclic quadrilateral with $ AB=BC=1$ and $ \angle ABC=120^{\circ}$. Then point $ D$ lies on an arc of a circle $ ABC$ and is no more distant from line $ AC$ then the midpoint of this arc. It is easy to see that in this case area$ (ABCD)=\sqrt{3}$. But initial pentagon was non-degenerated, hence this maximum is not attained. Consider the first case. We have $ DE=DC$. Consider a point $ P$, symmetric to $ E$ in $ AD$ and to $ C$ in $ DB$ (easy to see that it's the same point). Then $ P$ is a circumcentre of $ EDC$. Let rays $ EA$ and $ CB$ meet in $ Q$. Then we have area$ (ABCDE)$=area$ (QEDC)-\sqrt{3}/4$. Next, area$ (QEDC)$=area$ (QEC)$+$ EC^2\sqrt{3}/4$. Denote $ AE=AP=a$, $ BC=BP=b$. Then we get $ EQ=a+1$, $ QC=b+1$ and we have to prove (using cosine theorem and expression for area of triangle via two sides and angle $ 60^{\circ}$ between them): \[ \frac{\sqrt{3}}4\left((a+1)(b+1)+[(a+1)^2+(b+1)^2-(a+1)(b+1)]-1\right)< \sqrt{3}.\] It is equivalent to $ a^2+b^2+2(a+b)< 3$. But we have $ 3=1+2(a^2+b^2+ab)$ (as $ \angle APB=120^{\circ}$), substitute it to get equivalent obvious inequality $ (a+b-1)^2>0$.

Fedor Petrov wrote: ... Then $ P$ is a circumcentre of $ EDC$. .. Why?