Ireland_2017

Let $X$, $Y$ and $Z$ be the reflections in $BC$, $CA$ and $AB$ respectively and let $E$ and $F$ be the midpoints of $CA$ and $AB$ respectively.



Since $Z$ is the reflection of $O$ in $AB$, $AZ=AO=OB$. Similarly $AY=OC$. Also $ZZAB=LOAB$ and $ZYAC=LOAC$, hence $LBAC=\frac{1}{2}\angle ZAY$. Since $O$ is the circumcentre, $LBAC=\frac{1}{2}\angle BOC$ and so $\angle ZAY=\angle BOC$ which implies that $\angle ZAY$ is congruent to $\angle BOC$, hence $ZY=BC$. Now $ZB=OB=OC=CY$, hence $ZYCB$ is a parallelogram and $YZ\parallel BC$. Similarly $XY\parallel AB$ and $XZ\parallel AC$, and so the triangles $ABC$ and $XYZ$ are congruent and corresponding sides are parallel.

---

Alternative solution.

Let $X$, $Y$ and $Z$ be the reflections in $BC$, $CA$ and $AB$ respectively and let $E$ and $F$ be the midpoints of $CA$ and $AB$ respectively.



Because $E$ and $F$ are the mid-points of the sides $CA$ and $AB$, the Intercept Theorem (or the Mid-Point Theorem) implies that $EF\parallel BC$ and $|BC|=2|EF|$. Because $F$ is the mid-point of $OZ$ and $E$ the mid-point of $OY$, the same reason gives $EF\parallel YZ$ and $|YZ|=2|EF|$. Hence $YZ\parallel BC$ and $|YZ|=|BC|.$ Similarly, $XY\parallel AB$, $|XY|=|AB|$ and $XZ\parallel AC$, $|XZ|=|AC|$ and so the triangles $ABC$ and $XYZ$ are congruent and corresponding sides are parallel.