Romanian Mathematical Olympiad

Because $\triangle ABM \equiv \triangle ADN$ and $\triangle DAE \equiv \triangle DCF$, we have $m(\overline{BAM}) = m(\overline{DAN}) = m(\overline{ADE}) = m(\overline{CDF}) = 15^\circ$.
Then $\triangle$ ABM $\equiv \triangle$ ADN $\equiv \triangle$ DAE $\equiv \triangle$ DCF and $\triangle AMN \equiv \triangle DEF$.
It is clear now that $ADNE$ is a rectangle, and thus $P$ is the common midpoint of the line segments $[AN]$ and $[DE]$. Then, in the equilateral triangle $DEF$ we have $PF \perp DE$.
As $m(\overline{DAM}) + m(\overline{ADE}) = m(\overline{DAM}) + m(\overline{BAM}) = 90^\circ$, we get that $AM \perp DE$, obtaining that $AM \parallel PF$.
The equality $m(\overline{PAQ}) = m(\overline{PEQ}) = 60^\circ$ implies that $PAEQ$ is a cyclic quadrilateral, and thus $m(\overline{FPQ}) = m(\overline{EPF}) - m(\overline{EPQ}) = 75^\circ$.
On the other hand, $m(\overline{MFP}) = 180^\circ - m(\overline{CFD}) - m(\overline{DFP}) = 75^\circ$, which implies that $MQPF$ is an isosceles trapezoid and thus $[PQ] = [FM]$.