Let $ABCD$ be a convex quadrilateral, and let $N$ be the midpoint of $BC$. Suppose further that $\angle AND=135^{\circ}$. Prove that $|AB|+|CD|+\frac{1}{\sqrt{2}}\cdot |BC|\ge |AD|.$
Source: Baltic Way 2001
Tags: inequalities, geometry proposed, geometry
Let $ABCD$ be a convex quadrilateral, and let $N$ be the midpoint of $BC$. Suppose further that $\angle AND=135^{\circ}$. Prove that $|AB|+|CD|+\frac{1}{\sqrt{2}}\cdot |BC|\ge |AD|.$