In the acute-angled triangle $ABC$ all angles are greater than $45^o$. Let $AM$ and $BN$ be the heights of this triangle and let $X$ and $Y$ be the points on $MA$ and $NB$, respecively, such that $|MX| =|MB|$ and $|NY| =|NA|$. Prove that $MN$ and $XY$ are parallel.