THE 68th ROMANIAN MATHEMATICAL OLYMPIAD

Because $△ABM\equiv △ADN$ and $△DAE\equiv △DCF$, we have $m(\overline{BAM})=m(\overline{DAN})=m(\overline{ADE})=m(\overline{CDF})=15^{\circ }$. Then $△ABM\equiv △ADN\equiv △DAE\equiv △DCF$ and $△AMN\equiv △DEF$. 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 $[PQ]=[FM]$.