IRL_ABooklet

The quadrilateral CNOM is cyclic because $\angle CNO=90^{\circ }$ and $\angle CMO=90^{\circ }$, and $OC$ is a diameter and $L$ the centre of its circumcircle. The quadrilateral MNPQ is cyclic as well since $\angle PNQ=\angle PMQ=90^{\circ }$, and $PQ$ is a diameter and $K$ the centre of its circumcircle. The line MN is the radical axis of these two circles and so is perpendicular to the line KL which connects the centres of the circles. Because M and N are midpoints of sides of $△ABC$, $MN\parallel AB$, hence $AB\perp KL$.