In triangle $\vartriangle ABC$, $\angle BAC = 135^o$. $M$ is the midpoint of $BC$, and $N \ne M$ is on $BC$ such that $AN = AM$. The line $AM$ meets the circumcircle of $\vartriangle ABC$ at $D$. Point $E$ is chosen on segment $AN$ such that $AE = MD$. Show that $ME = BC$.