$AB=CD,AD \parallel BC$ and $AD>BC$. $\Omega$ is circumcircle of $ABCD$. Point $E$ is on $\Omega$ such that $BE \perp AD$. Prove that $AE+BC>DE$
Source: St Petersburg Olympiad 2015, Grade 9, P2
Tags: geometry, circumcircle, trapezoid
$AB=CD,AD \parallel BC$ and $AD>BC$. $\Omega$ is circumcircle of $ABCD$. Point $E$ is on $\Omega$ such that $BE \perp AD$. Prove that $AE+BC>DE$