75th NMO Selection Tests

From the condition $AC\perp BE$, it follows that $AC$ is the perpendicular bisector of segment $BE$, so $MB=ME$. Since $AB=AE$, we conclude that $△MAB\equiv △MAE$ (congruent triangles). It follows that $\angle MAB=\angle MAE$ (1), and $\angle MBA=\angle MEA$ (2). From $AB=AD$, it follows that triangle $ABD$ is isosceles with base $BD$, so $\angle ABD=\angle ADB$ (3). From (2) and (3), we deduce that $\angle MAE=\angle MDA$, so quadrilateral $MAED$ is cyclic. It follows that $\angle AED=\angle DMN$, and since $\angle AMB=\angle DMN$, we obtain $\angle AED=\angle AMB$ (4). From (1) and (4), it follows that $△MAB\sim △AEN$, so $\frac{AN}{AB}=\frac{AE}{AM}=2$, from which we conclude $AN=2\cdot AB$.