Maths Olympiad problem relating to circles? 

Given:
HG is the common tangent to circle C and J.
Prove:
$\angle{HFI} = \angle{HFG}$ or $HF$ bisects $\angle{GFI}$

My attempt:
$$\angle{FHG} + \angle {HGL} = \angle{LIF} + \angle {LFI}$$
$$\begin{align} \Rightarrow \angle{LFI}  = &\angle{FHG} + \angle {HGL} - \angle {LIF} \\  =   &\angle{IFG} + \angle {LGF} - \angle {HGF} - \angle {HFG}+ \angle {HGL} \\ = & \angle{IFG} + \angle {LGF} - \angle {LGF} - \angle {HFG} \\ = & \angle{IFG} - \angle {HFG} \end{align}$$
$$\angle{LFI} + \angle {HFG} = \angle{IFG}$$
which is not of any interest to me. 
However I start, I end at $\angle{LFI} + \angle {HFG} = \angle{IFG}$
A hint or two might be helpful.
Thanks.
Note: this was a problem in an Olympiad problem where 3 hours are given for solving 10 question subjectively, so I don't think there is a trick answer to this question.


Since $HG$ is tangent to the circle $J$, 
$$\angle{GHF}=\angle{HIF}$$
Similarly, letting $M\ (\not=F)$ be the point both on $IF$ and on the circle $C$, 
$$\angle{HGF}=\angle{GMF}=\angle{FEI}=\angle{FHI}$$
Now, consider the triangles $\triangle{FGH},\triangle{HFI}$ to get the result.
