International Mathematical Olympiad 2018 - Problem 1 
Let $\Gamma$ be the circumcircle of an acute-angled triangle $ABC$.  Points $D$ and $E$ lie on segments $AB$ and $AC$, respectively, such that $AD = AE$. The perpendicular bisectors of $BD$ and $CE$ intersect the minor arcs $AB$ and $AC$ of $\Gamma$ at points $F$ and $G$, respectively.  Prove that the lines $DE$ and $FG$ are parallel (or are the same line).

 A: Let $O$ be the center of $\Gamma$.  The perpendicular bisectors of the sides $AB$ and $AC$ meet the minor arcs $AB$ and $AC$ at $P$ and $Q$, respectively.  Suppose that the perpendicular bisectors of the line segments $BD$ and $CF$ intersect at $M$.
Note that $DE$ is perpendicular to the (internal) angular bisector $\ell$ of $\angle BAC$.  As $$\angle OPQ=\angle OQP=\frac{1}{2}\angle BAC\,,$$ it follows immediately that $\ell \perp PQ$.  This means $DE\parallel PQ$.
The distance $d_1$ between the parallel lines $OP$ and $MF$ equals $\frac{1}{2} AD$; likewise, the distance $d_2$ between the parallel lines $OQ$ and $MG$ is $\frac{1}{2} AE$.  Because $AD=AE$, we conclude $d_1=d_2$.  
Now, rotate the line $OP$ about the point $O$ until this line coincides with the line $OQ$ in such a way that the image of the line $MF$ under this rotation is exactly the line $MG$ (this is possible because $d_1=d_2$).    Let $P'$ and $F'$ be the images of $P$ and $F$, respectively, under this rotation.  Therefore, $P'QGF'$ is a cyclic quadrilateral with parallel opposite sides $P'Q$ and $F'G$.  Thus, $PF=P'F'=QG$, so $PFGQ$ is a cyclic quadrilateral with two opposite sides $PF$ and $QG$ having equal length.  This means $PQ \parallel FG$, and we can then conclude that $DE\parallel FG$, as required.

