Problem 1185

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.

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.