Problem

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Tags: geometry, trigonometry, inequalities, cyclic quadrilateral, geometry proposed



In a convex pentagon $ ABCDE$, let $ \angle EAB = \angle ABC = 120^{\circ}$, $ \angle ADB = 30^{\circ}$ and $ \angle CDE = 60^{\circ}$. Let $ AB = 1$. Prove that the area of the pentagon is less than $ \sqrt {3}$.