74th NMO Selection Tests for JBMO

a) Construct the equilateral triangle $BDQ$, such that $C$ lies in its interior. $Q$ is situated on the perpendicular bisector of the diagonal $BD$, therefore $Q$, $C$ and $A$ are collinear. Since $\angle EBQ=\angle FBD=60^{\circ }-\angle FBQ$, $BQ=BD$ and $BE=BF$, triangles $BEQ$ and $BFD$ are congruent (S.A.S.), thus $\angle BQE=\angle BDF=\angle ADF-\angle ADB=15^{\circ }$ and $QE=DF=DM=AM$. From $\angle QBC=\angle QBD-\angle CBD=15^{\circ }=\angle BQE$, we infer that $QE\parallel CB\parallel AD$. As $QE=AM$, it follows that $AMQE$ is parallelogram, so the midpoint $P$ of $ME$ lies on $AQ$, i.e., $P$ is situated on the line $AC$.

b) Since $QE=DM$ and $QE\parallel DM$ we deduce that $DMEQ$ is a parallelogram and $\angle DME=\angle DQE=75^{\circ }$. Consequently, $\angle FMP=\angle DMP-\angle DMF=15^{\circ }$. The triangle $FAD$ is right angled, with $\angle FAD=30^{\circ }$, therefore we obtain $\angle FAP=\angle DAP-\angle DAF=15^{\circ }$ and $\angle MFA=\angle DFA-\angle DFM=30^{\circ }$. Since $\angle FMP=\angle FAP$, it follows that $AMFP$ is a cyclic quadrilateral, with $\angle MPA=\angle MFA=30^{\circ }$. $\angle FPM=\angle FAM=30^{\circ }$, therefore $\angle FPM=\angle APM$, and $PM$ is the angle bisector of $APF$.