Maths Olympiad problem relating to circles?

Question

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Given:

HG is the common tangent to circle C and J.

Prove:

$\angle{HFI} = \angle{HFG}$ or $HF$ bisects $\angle{GFI}$

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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.

Answer 1

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.

Answer 2

Direct one-line proof: $\angle GFH = \angle GFE + \angle EFH = \angle HGE + \angle EHG = \angle HEI = \angle HFI$.

Nothing extra constructed!