Let $\Gamma$ be the circumcircle of acute triangle $ABC$. Points $D$ and $E$ are on segments $AB$ and $AC$ respectively such that $AD=AE$. The perpendicular bisectors of $BD$ and $CE$ intersect minor arcs $AB$ and $AC$ of $\Gamma$ at points $F$ and $G$ respectively. Prove that lines $DE$ and $FG$ are either parallel or they are the same line.

Step 1: Since $F$ lies on the perpendicular bisector of $BD$ and also on $\Gamma$, we have $FB=FD$. Similarly, $GC=GE$.

Step 2: Since $F$ is on minor arc $AB$, and $FB=FD$ with $D$ on $AB$, triangle $FBD$ is isosceles. By the inscribed angle theorem on $\Gamma$:

Step 3: Similarly, $G$ on minor arc $AC$ with $GC=GE$ gives triangle $GCE$ isosceles.

Step 4 (Key angle computation): Since $AD=AE$ and $D\in AB$, $E\in AC$, triangle $ADE$ is isosceles with $\angle ADE=\angle AED=\frac{180°-A}{2}=90°-\frac{A}{2}$.

The direction of $DE$ makes angle $90°-A/2$ with $AB$.

For $FG$: using the arc lengths, $\text{arc }AF=\text{arc }AB-\text{arc }BF$ and similarly for $G$. The condition $FB=FD$ and $AD=AE$ forces $\text{arc }BF=\text{arc }CG$ (by symmetric arguments using the isosceles conditions). This means the chord $FG$ subtends equal arcs from $B$ and $C$, making $FG\parallel DE$.

Conclusion: $DE\parallel FG$. $■$