$\boxed{\text{Solution 1.}}$ Consider point Z on AB in such a way that $AZFD$ is an isosceles trapezoid hence it is cyclic. Let $O$ be the circumcenter of $\triangle {AFD}$. Since $X$ and $Y$ are the reflections of $E$ and $C$ with respect to $AD$ and $AF$ , and $ZBCF$ is a parallelogram, then $ED = XD = CF = F Y = ZB$ Suppose that $AF$ meets $CY$ at $H.$ Now notice that $\angle OZB = \angle OZF + \angle FZB = 90^{\circ} -\angle ZDF + \angle BCD = 90^{\circ} - \angle AFD + \angle ADC$ $\angle ODX = \angle ODA + \angle ADX = 90^{\circ} -\angle AFD + \angle ADC$ $\angle OFY = 180^{\circ} - \angle OF A - \angle YFH = 180^{\circ} - (90^{\circ} - \angle ADC) - \angle HFC = 90^{\circ} - \angle AFD + \angle ADC$ These three equality with $ZB = XD = FY$ and $OZ = OD = OF$ gives us that $ \triangle OZB \cong \triangle ODX \cong \triangle OFY \Longrightarrow OB = OX = OY $ We are done!