Just a bit of formalization (nothing more towards a proof). The harmonic mean $h$ of $AB$ and $CD$ is $h = \dfrac {2}{\dfrac {1}{AB} +\dfrac {1}{CD}} = \dfrac {2AB\cdot CD}{AB + CD} = \dfrac {2g^2}{s}$, where $g$ is the geometric mean of $AB$ and $CD$, and $s = AB+CD = BC + DA$ is the semiperimeter of the quadrilateral (by Pitot's theorem). Then $\bullet$ If $KL = \dfrac {2d^2}{s}$, then $(KL=h) \implies (g=d)$, while $(g=d) \implies (KL=h)$, therefore $(KL = h) \iff (g=d)$; $\bullet$ If $KL \neq \dfrac {2d^2}{s}$, then $(KL=h) \implies \lnot(g=d)$, while $(g=d) \implies \lnot(KL=h)$ (but this is just the first one), therefore $(KL = h) \land (g=d) = 0$. This is not quite enough; we need show that one (any) of $(KL = h) = 1$ and $(g=d) = 1$ never holds.