62nd Czech and Slovak Mathematical Olympiad

Alternate angles $ABD$ and $CDB$ are equal (Fig. 2), hence $\angle BCA+\angle ABD+\angle BDA+\angle ACD=180^{\circ }$. The equality $\angle BCA+\angle ABD=\angle BDA+\angle ACD$ thus holds if and only if $$\angle BCA+\angle ABD=90^{\circ }.$$ (1) Fig. 2 Points $K$ and $L$ lie on a circle with diameter $BD$. Hence the inscribed angles $BDK$ and $BLK$ are equal and (due to equal alternate angles $ABD$ and $CDB$) $$\angle BLK+\angle ABD=\angle BDK+\angle CDB=90^{\circ }.$$ Lines $KL$ and $AC$ are parallel if and only if $\angle BLK=\angle BCA$ which is by the last equality equivalent to (1). The equivalence is thus proven.