Indian National Mathematical Olympiad

We have $FG = FA$ since $FG$ is parallel to $BC$. But also $\triangle GAE$ is a right angle triangle. Thus, if $F'$ is the midpoint of $GE$, then $\angle GAF = \angle FGA = \angle F'GA = \angle GAF'$ which implies $F \equiv F'$. Thus, $F$ is the midpoint of $GE$.

If $O$ is the circumcenter of $\triangle EBG$, then
$$\angle FOE = \angle GBE = \angle ABE = \angle ACE = \angle FCE.$$
Thus, we get $\angle FOE = \angle FCE$ as desired. $\square$

($\angle BCA$ acute case) Let $O_1$ be the circumcenter of $\triangle ABC$, $O$ be the circumcenter of $\triangle EBG$ and $\omega$ be the circumcircle of $\triangle ECF$.

First, show that $F$ is the midpoint of $EG$ as in Solution 1. Next, we show that $O_1$ lies on $\omega$. This follows from
$$\angle EO_1C = 2\angle EBC = 2\angle O_1BC = 2\angle BCO_1 = \angle BCA = \angle EFC.$$
Now, $O_1$ is the midpoint of $EB$ and $F$ is the midpoint of $EG$, therefore the homothety at $E$ with ratio $1/2$ takes $\triangle EGB$ to $\triangle EFO_1$. Thus, it takes $O$, the circumcenter of $\triangle EGB$, to the circumcenter of $\triangle EFO_1$, thus proving that the midpoint of $EO$ is the center of $\omega$. This immediately implies that $O$ lies on $\omega$. $\square$

($\angle BCA$ acute case) Let $O_1$ be the circumcenter of $\triangle ABC$, $O$ be the circumcenter of $\triangle EBG$ and $\omega$ be the circumcircle of $ECF$.
First, show that $F$ is the midpoint of $EG$ as in Solution 1. Next, we show that $O_1$ lies on $\omega$. This follows from
$$\angle EO_1C = 2\angle EBC = 2\angle O_1BC = 2\angle BCO_1 = \angle BCA = \angle EFC.$$
Now, $O_1$ is the midpoint of $EB$ and $F$ is the midpoint of $EG$, therefore the homothety at $E$ with ratio $1/2$ takes $\triangle EGB$ to $\triangle EFO_1$. Thus, it takes $O$, the circumcenter of $\triangle EGB$, to the circumcenter of $\triangle EFO_1$, thus proving that the midpoint of $EO$ is the center of $\omega$. This immediately implies that $O$ lies on $\omega$. $\square$