过点O作OE⊥AB于E∵AB∥CD，∴OE⊥CD于F∵AC=BD，∠ACD=∠BDC，CD=DC∴△ACD≌△BDC．∴∠ADC=∠BCD又∵BC⊥AD，∴∠ADC=∠BCD=45°∴OF=12CD同理可得，OE=12AB∴EF=12（AB+CD）又∵中位线=12（AB+CD）=a∴S梯形ABCD...