Taiwan IMO Selection Camp

Extend $BC$ to meet the common tangent line of the two circles at $R$. Extend $TE$ to meet circle $O$ again at a second point $H$.
First, since $TR$ is the common tangent line of the two circles, we have $\angle GHT = \angle RTF = \angle FET$, hence $EF \parallel GH$.
Next, note that $\angle RTC = \angle TBC$, and since $RF$ and $RT$ are both tangent lines to circle $K$, we have
$$\angle RTF = \angle RFT = \angle TBC + \angle BTG = \angle RTC + \angle BTG,$$
$$\angle BTG = \angle RTF - \angle RTC = \angle CTG.$$
Thus $G$ is the midpoint of arc $BC$.

(1) From $EF \parallel GH$ we know: $\angle AIE = \angle AGH$. Also, $\angle AGH = \angle ATH$, and hence $\angle AIE = \angle ATH$. Therefore $A$, $E$, $T$, $I$ are concyclic.

(2) First, since $A$, $E$, $T$, $I$ are concyclic, $\angle AIT = \angle AET = \angle EFT$, so the circumcircle of $\triangle FTI$ is tangent to $AG$ at $I$.
Next, since $G$ is the midpoint of arc $BC$, we have $\angle BAG = \angle CAG$ and $\angle BTG = \angle CTG$. Also $\angle GBC = \angle GAC$ and $\angle BTG = \angle BAG$, so $\angle GBC = \angle BTG$. Therefore, the circumcircle of $\triangle BFT$ is tangent to $GB$ at $B$.
(3) Finally, since $GT$ is a chord from $G$ to the circumcircle of $\triangle FTI$ and the circumcircle of $\triangle FTI$, while $GB$, $GI$ are tangent lines to these two circles respectively, we have $GB^2 = GF \cdots GT = GI^2$, that is, $GB = GI$. Q.E.D.!