67th NMO Selection Tests for JBMO

Let $I_a$ be the center of the circle $k$ and let $T$ be the midpoint of the arc $BC$ - not containing $A$ - of the circumcircle $ABC$. Notice that $OT$ is the perpendicular bisector of the line segment $BC$ to deduce that $OT\perp BC$. As $I_{a}D\perp BC$ and $OT=I_{a}D=R$, the quadrangle $OD{I}_{a}T$ is a parallelogram. Consequently $OD\parallel T{I}_a$, or, equivalently, $OD\parallel A{I}_a$.

On the other hand, since $AE=AF$ and $A{I}_a$ bisects angle $\angle FAE$, the lines $A{I}_a$ and $EF$ are perpendicular, and so are lines $OD$ and $EF$.