If you square each side, move the denominators to the numerators on the other side, bring everything over to the left, and carry out the multiplication, it looks like you will get quartic equation for $F$. A closed form solution for such equation an exist, though it gets a bit messy.
Prior to carrying out the multiplication, it will look like this:
(aF-b)^2 \left( (hF-i)^2 + (jF-k)^2 +(mf-n)^2 \right) - (hF-i)^2 \left( (aF-b)^2 + (cF-d)^2 +(ef-g)^2 \right) = 0
Eventually you will end up with something of the form $\alpha F^4 + \beta F^3 + \gamma F^2 + \delta F + \epsilon = 0$, the solutions to which can be found at the linked Wikipedia page.