62nd Belarusian Mathematical Olympiad

If the quadrilateral $ABCD$ has the pair of parallel sides the result follows from the symmetry of the construction. So we suppose that $ABCD$ is different from a trapezoid and a rectangle. Let $X=BC\cap AD$. Since ${\Gamma }_2$ touches $AD$ at point $F$, we have $X{F}^2=XB\cdot XC$. Similarly, $X{H}^2=XA\cdot XD$. Since $ABCD$ is cyclic, we have $XB\cdot XC=XA\cdot XD$, so $XF=XH$, and hence $△XHF$ is an isosceles triangle. Thus $\angle BHF=0.5(180^{\circ }-\angle CXD)=0.5(\angle C+\angle D)$. Similarly, $\angle BGE=0.5(\angle A+\angle D)$. Let $GE\cap HF=Y$. Then $△GYH=360^{\circ }-(\angle B+0.5(\angle C+\angle D))+0.5(\angle A+\angle D))=360^{\circ }-(\angle B+\angle D)-0.5(\angle A+\angle C)$. Since $\angle A+\angle C=\angle B+\angle D=180^{\circ }$, we have $△GYH=360^{\circ }-180^{\circ }-90^{\circ }=90^{\circ }$, so $EG\perp FH$.