Let D' be reflection of D about K. Note that $S_{DOK}$ = $S_{D'OK}$ so we need to prove $S_{AOB}$ = $S_{AOD'}$. We will show D' and B are equidistant from AC. Note that B' is reflection of B about AC so we need to prove B'D' || AC. Let S be midpoint of BC. Note that B'D || BO and SO || B'C || B'K and KD || BS so B'KD and OSB are homogeneous. Also note that DK/DD' = BS/BC so DB'D' and BOC are homogeneous as well. now we have ∠B'D'D = ∠OCB = ∠OAD so AC || B'D' as wanted. we're Done.