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 $APF$.