Proof. Notice that \angle{BDC}=\angle{BAC}, so
\angle{GBI}=\frac{\angle{ABC}}{2}-(\angle{DBG}-\angle{DBA})=
\angle{GBI}=\frac{\angle{ABC}}{2}-(90^\circ-\frac{\angle{BAC}}{2})+\angle{DCA}.
Replacing 90^\circ by \frac{\angle{ABC}}{2}+\frac{\angle{BAC}}{2}+\frac{\angle{BCA}}{2} and simplifying, we get \angle{GBI}=\angle{DCA}-\frac{\angle{BCA}}{2}. But \angle{DCA}=\angle{BCA}-\angle{BCD}, thus, \angle{GBI}=\frac{\angle{BCA}}{2}-\angle{BCD}. Moreover, \angle{DCI}=\angle{GCI}=\frac{\angle{BCA}}{2}-\angle{BCD}, hence,\angle{GBI}=\angle{GCI} implying BGIC is cyclic.
\square
Lemma 2. IE=IG.
Proof. Notice that \angle{BEG}=\frac{\angle{BDG}}{2}=\frac{\angle{BAC}}{2}=\angle{BAI} which means that EG\parallel{AI} implying that ID\perp{EG} and as DI passes through the center of circle (D, DB) we deduce DI is the perpendicular bisector of EG meaning IE=IG.
\square
Back to our main problem.
\angle{IEG}=\angle{IGE}=360^\circ-\angle{EGF}-\angle{DGB}-\angle{BGC}-\angle{CGI}.
Combining lemmas 1 and 2 we get \angle{EGF}=\frac{\angle{BAC}}{2}+\angle{DCA}; \angle{DGB}=90^\circ-\frac{\angle{BAC}}{2}; \angle{BGC}=\angle{BIC}=180^\circ-\frac{\angle{ABC}}{2}-\frac{\angle{BCA}}{2}; \angle{CGI}=\frac{\angle{ABC}}{2}.
Replacing and simplifying we get \angle{IEG}=90^\circ+\angle{BCD}-\frac{\angle{BCA}}{2}. Finally, \angle{FEI}+\angle{FCI}=180^\circ+\angle{BCD}-\frac{\angle{BCA}}{2}+\frac{\angle{BCA}}{2}-\angle{BCD}=180^\circ and we are done.
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