设P在AB上,Q在CD上,M在BC上,N在AD上,且PQ=MN.过A作AE‖PQ交CD于E,过D作DF‖MN交BC于F,∴AE=PQ,DF=MN,得AE=DF,由AD=CD,∴△ADE≌△DCF（H,L）∴∠DAE=∠CDF,又∠DAE+∠DEA=90°,∴∠CDF+∠DEA=90°,∴AE⊥DF,∴PQ⊥MN...