Croatian Mathematical Olympiad

We easily see that $\angle AKE=\angle KEA=\angle DEC=\angle CDE$. Thus $|AK|=|AE|$ and $AK\parallel CD$.



Let $L^{\prime }$ be the intersection of the lines $CP$ and $AK$. Since the triangles $PBC$ and $PA{L}^{\prime }$ are similar, by trigonometric calculus we get $|BP|:|AP|=|BC|:|A{L}^{\prime }|$, from which it follows that $|A{L}^{\prime }|=|AK|=|AE|$, i.e. $L^{\prime }$ lies on the circle $k$ - so it must be $L^{\prime }\equiv L$. Similarly, we can show that $D,F$ and $L$ are collinear.

Now we know that $\overline{KL}$ is the diameter of the circle $k$. Observing the triangles $AKO$ and $EAI$, we notice they are both right-angled, and $\angle AKO=90^{\circ }-\angle KOA=90^{\circ }-\frac{1}{2}\angle KOL=90^{\circ }-\angle KDL=\angle IAE$ holds. Hence they are congruent, which means that segments $\overline{AO}$ and $\overline{ID}$ are parallel (since they are perpendicular to $KL$ and $BC$, respectively) and have equal length ($|ID|=|IE|=|OA|$). Therefore, the quadrilateral $AODI$ is a parallelogram, from which it follows that $AI\parallel OD$.