The 14th Thailand Mathematical Olympiad

Suppose that triangle $ABC$ is isosceles. Since $\angle ACB=45^{\circ }$ and $ABC$ is an acute triangle, we must have $AC=BC$.



Since $CM$ is the perpendicular bisector of $AB$, $\angle BCM=\frac{1}{2}\angle BCA=22.5^{\circ }$. Since $$\angle BAD=\angle BAC-\angle DAC=22.5^{\circ }=\angle BCM.$$ Hence triangles $BAD$ and $NCD$ are congruent (two angles and $AD=CD$), so $CN=AB=2AM$ as required.

Conversely, suppose $CN=2AM$. Since $CD=AD$ and $CN=2AM=AB$, so triangles $CDN$ and $ADB$ are congruent. Thus, $\angle NCD=\angle BAD$. Since $\angle ANM=\angle CND$ and $\angle MAN=\angle NCD$, we have $\angle AMN=\angle CDN=90^{\circ }$. In addition, $AM=MB$, and so triangles $AMC$ and $BMC$ are congruent. Thus, $AC=BC$.