75th Romanian Mathematical Olympiad

$\angle GBF=\angle DBC=45^{\circ }$, thus the triangle $BDF$ is right-angled at $B$. Since $EG$ is the perpendicular bisector of $BF$ in triangle $BDF$, it follows that $H$ is the midpoint of the hypotenuse $DF$.

Let $K$ be the projection of $H$ onto $AE$. Since $H$ is the midpoint of $DF$ and $HK\parallel AD$, it follows that $HK$ is the midsegment in the right trapezoid $AEFD$, therefore $K$ is the midpoint of the side $AE$. Consequently, the triangle $HAE$ is isosceles, so $\angle HAE=\angle HEA=45^{\circ }$, thus $H\in AC$.

Let $L$ be the projection of $H$ onto $AD$. We have $△HAK\equiv △HAL$ (HA), therefore $HK=HL$. Since $\angle KHI+\angle IHL=90^{\circ }=\angle DHL+\angle IHL$, we obtain $\angle KHI=\angle LHD$, hence $△HKI\equiv △HLD$ (LA), therefore $HI=HD$.

Since $△HAB\equiv △HAD$ (SAS), we have $HB=HD=HI$. Consequently, $K$ is the midpoint of $BI$, $KH$ is a midsegment in the triangle $IBJ$ and $H$ is the midpoint of $IJ$. Thus, the diagonals of the quadrilateral $DIFJ$ are equal, bisect each other and are orthogonal, therefore $DIFJ$ is a square.

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Alternative solution.

As in the first solution, $H$ is the midpoint of $DF$. Since $IJ$ is the perpendicular bisector of the segment $DF$, we have $DI=IF$. Moreover, $\angle IHF=\angle IEF=90^{\circ }$, thus $IEFH$ is a cyclic quadrilateral, therefore $\angle IFH=\angle IEH=45^{\circ }$. Since the triangle $IFD$ is isosceles, we have $\angle IFD=\angle IDF=45^{\circ }$ and $\angle DIF=90^{\circ }$.

Let $K$ be the projection of $H$ onto $AE$. From $\angle AID+\angle EIF=90^{\circ }$ we deduce that $\angle AID=\angle EFI$, hence $△AID\equiv △EFI$ (HA), therefore $AI=EF=BE$. Thus, $K$ is the midpoint of $IB$. Since $HK\parallel BJ$, it follows that $HK$ is a midsegment of the triangle $BIJ$, so $H$ is also the midpoint of $IJ$. Since its diagonals $DF$ and $IJ$ bisect each other, it follows that $DIFJ$ is a square.