In a convex pentagon $ ABCDE$ side $ AB$ is perpendicular to $ CD$ and side $ BC$ is perpendicular to $ DE$. Prove that if $ AB = AE = ED = 1$, then $ BC + CD < 1$.
Source: Russia 1994
Tags: inequalities, geometry proposed, geometry
In a convex pentagon $ ABCDE$ side $ AB$ is perpendicular to $ CD$ and side $ BC$ is perpendicular to $ DE$. Prove that if $ AB = AE = ED = 1$, then $ BC + CD < 1$.