Very easy for a $P6$! Area of $ABC$ equals to the sum of areas of triangles $BFC$ and $AFC$. So $2A_{ABC}=CF\cdot (a+b)\cdot sin\frac{\gamma }{2}=AD\cdot BC=BE\cdot AC.$ $(1)$ Condition $ \frac{CF}{AD}+\frac{CF}{BE}=2 $ is equivalent to $CF\cdot (A+B)=4A_{ABC}.$ $(2)$. From $(1)$ and $(2)$ we conclude that $\angle ACB=\gamma =\frac{\pi }{3}$. It is now easy to prove that $CO=CH$,so line $CI$ is a perpendicular bisector of the segment $OH$, thus $IO=IH$.

yea ,easy for a IMO TST! proof = note that by sine law in triangle $CBF$ we get $FC=\frac{2RsinAsinBsinC}{(sinA+sinB)sinC/2} $ also note that $BE=2RsinAsinC$ and $AD=2RsinBsinC$ thus putting above expressions in this $\frac{CF}{AD}+ \frac{CF}{BE}=2$ yield $\angle C=60$ and now it becomes trivial by some angle chase. first note that we get $A,B,H,I,O,J$ are concyclic where $J$ is $C$-excenter. with diameter $IJ$ now concyclicity of these points yield $\angle HBJ=180-A-B/2$ and $\angle OAJ=120-A/2$ and thus $\angle HBJ=\angle OAJ$ now this implies $\angle OIJ=\angle HIJ$ or $\angle HJI=\angle IJO$ or $\angle HOI=\angle OHI$ and hence $IH=IO$ we are done