67th Czech and Slovak Mathematical Olympiad

Let $K$ be the incenter of triangle $ABD$. Since $IK\parallel AB$, it suffices to show $JK\parallel AB$. Let $\angle ABD=\angle ACD=\phi$. Then $\angle AKD=90^{\circ }+\frac{1}{2}\phi$ and $\angle DJA=90^{\circ }-\frac{1}{2}\phi$, implying that the quadrilateral $AKDJ$ is cyclic (Fig. 3).

Fig. 3

As $AK$, $DJ$ are bisectors of alternate interior angles, they are parallel. Together with the cyclic quadrilateral we obtain $\angle AKJ=\angle ADJ=\angle DAK=\angle KAB$ which concludes the proof.