Another Tran Viet Hung's problem

Lemma 1. Let $CD$ cuts circle $(D, DB)$ in $G$. Then, $BGIC$ is cyclic.

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.