Proceed with complex numbers. Fix $a, b, d$ on the unit circle, and since $AC \perp BD$ we get $ac + bd = 0 \implies c = \tfrac{-bd}{a}$. Let $K$ be the reflection of $C$ in $BD$, we are required to prove that $K \equiv F$. Compute $k = \tfrac{-bd}{a} (b + d + abd)$. Assign complex number $f$ to point $F$ and $g = \overline{f}$ for convenience. Since $ A, F, C$ are collinear, by the collinearity criterion, $$\tfrac{f-a}{f+\tfrac{bd}{a}} = \tfrac{g-\tfrac{1}{a}}{g + \tfrac{a}{bd}}$$Since $DF \perp AB$, by the perpendicularity criterion, $$\tfrac{d-f}{b-a} + \tfrac{\tfrac{1}{d} - g}{\tfrac{1}{b} - \tfrac{1}{a}} = 0$$Solving the equations keeping $f, g$ as variables gives our desired answer that $f = \tfrac{-bd}{a}(b + d + abd) = k \implies FB = BC$.

Let $L=AC \cap BD$ Notice that: $\angle ACB=\angle ADB=\angle AEL$ by cyclic Now , as $FELB$ is cylic $\angle AEL=\angle LFB$ $\implies FB=BC$