Romanian Mathematical Olympiad

a) We denote $\angle BAE = \angle AEM = \angle AMP = \alpha$.
We have, in turn $\angle DAE = 90^\circ - \alpha$, $\angle EAC = \alpha - 45^\circ$, $\angle DEA = \angle EAB = \alpha$, $\angle CEM = 180^\circ - 2\alpha$.
From the triangle $CPM$ follows that $\angle PMC = 180^\circ - \angle AMP = 180^\circ - \alpha$ and then $\angle CPM = \alpha - 45^\circ$. Thus $\angle EAC = \angle CPM = \alpha - 45^\circ$.
Denote by $H$ the intersection of the lines $AE$ and $PM$. The quadrilateral $HAPC$ has $\angle HAC = \angle HPC = \alpha - 45^\circ$ so it is inscribed, hence $\angle AHP = \angle ACP = 45^\circ$. Moreover, $\angle AHP = \angle EHM = \angle ECM = 45^\circ$, so the quadrilateral $HEMC$ is inscribed and $\angle HCM = \angle AEM = \alpha$.
Returning to the inscribed quadrilateral $HAPC$, we have $\angle HPA = \angle HCA = \alpha$, so the triangle $AMP$ is isosceles.

b) In the isosceles triangle $MAP$ we have $\angle MAP = 180^\circ - 2\alpha$, thus $\angle PAB = \angle CAB - \angle MAP = 45^\circ - (180^\circ - 2\alpha) = 2\alpha - 135^\circ$.
We extend side $CE$ with segment $DQ = BP$. From the congruence of the triangles $ADQ$ and $ABP$ (SAS) it follows that $\angle QAD = \angle PAB = 2\alpha - 135^\circ$. Thus $\angle EAQ = \angle EAD + \angle DAQ = \alpha - 45^\circ$, which means $\angle EAQ = \angle EAM$. Now the congruence of the triangles $EAQ$ and $EAM$ (ASA) yields $EM = EQ = ED + DQ = ED + BP$.