Note that $D$ is the reflection of the orthocenter of $ABC$ over $AC$, so $CD = CH' = 2OH$, where $H'$ is the orthocenter of $ABC$.

Using complex numbers: Set $a,b,c,d$ on the unit circle with $d=\frac{-ac}{b}$ This gives us: \[ \left\vert OF \right\vert ^2 = \left\vert \frac{a+b}{2} \right\vert ^2 = \frac{(a+b)(\frac{1}{a}+\frac{1}{b})}{4} = \frac{(a+b)^2}{4ab} = \frac{(c-d)(\frac{1}{c}-\frac{1}{d})}{4} = \left \vert \frac{c-d}{2} \right \vert^2 \]