Selection tests for the Balkan Mathematical Olympiad 2013

Because $AEMD$ is cyclic, we have $\angle BEM=\angle ADM$. But $\angle MBE=\angle MAD=45^{\circ }$ and $BM=AM$. We deduce that triangles $BME$ and $AMD$ are congruent and therefore $ME=MD$.

Because $AEMD$ is cyclic, $\angle DME=90^{\circ }$. Therefore, the area of triangle $EMD$ is $$2=\frac{1}{2}ME\cdot MD=\frac{1}{2}M{D}^2$$ and hence, $MD=ME=2$.



We have from the cosine law in triangle $AMD$ $$4=M{D}^2=A{D}^2+A{M}^2-2AD\cdot AM\cos 45^{\circ }=A{D}^2+\frac{9}{2}-3AD$$ Because $AE$ satisfies the same equation as $AD$ above, and $AE<AD$, we deduce that $AD=\frac{3+\sqrt{7}}{2}$ and therefore $$CD=\frac{3-\sqrt{7}}{2}.$$