74th 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 $EM=EQ=ED+DQ=ED+BP$.