$ABDE, BCEF, CDFA$ are simetric trapezoids, since their diagonals are equal and two of their sides are parallel. Then $\angle ABE=\angle ADE=\angle DEB=\angle DAB$ and $\angle BCF=\angle BEF=\angle EFC=\angle EBC$ and $\angle CDA=\angle CFA=\angle FAD=\angle FCD$ Then $\angle ABE+\angle EBC +\angle CDA = 1/4*(\angle A+\angle B+\angle C+\angle D+\angle E+ \angle F) = \pi$ Then $ABCD$ is cyclic. $ABDE, CDFA$ are cycilc, then $E, F$ are in the circumcircle of $ABCD$ Then $ABCDEF$ can be inscribed in a circle

AB //ED, AD= BE so ABDE is either rectangle (# with equal diagonals) or isosceles trapezoid, in any case ABDE is cyclic similarly BCEF and CDFA are cyclic By the three cyclic quads we conclude that ABCDEF is cyclic