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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).

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closed as off-topic by Crostul, Namaste, rschwieb, user 170039, Wojowu Jul 12 '18 at 14:13

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    $\begingroup$ Is this an on-going contest ? $\endgroup$ – Peter Jul 12 '18 at 11:48
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    $\begingroup$ @Peter. The IMO has finished, as far as I know. $\endgroup$ – MalayTheDynamo Jul 12 '18 at 11:49
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    $\begingroup$ IMO is such a big deal that one can expect to find solutions online. I did a quick search and found, among other things, this video in a couple of seconds. I haven't watched it, so I can't vouch for correctness, but there are bound to be other materials out there. AoPS probably has a couple of pages on the problem as well, I'd wager. $\endgroup$ – Arthur Jul 12 '18 at 12:03
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    $\begingroup$ I don't see why IMO problems can't be posted here. There may be new solutions, and we should invite them. $\endgroup$ – Batominovski Jul 12 '18 at 14:19
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    $\begingroup$ @rschwieb Why? You think this problem is easy? It's probably the easiest of IMO2018, but it's not easy compared to most questions being asked on this site. I can sympathize with the OP, who might have gotten stuck without ideas. $\endgroup$ – Batominovski Jul 12 '18 at 17:38
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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.

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