证：∵ABCD为平行四边形，∴AD=BC,AD∥BC∵AD∥BC∴∠ADE=∠CBF.AE⊥BD CF⊥BD∠AED=∠CFB=90 且AE∥FC∵AD=BC ∠ADE=∠CBF ∠AED=∠CFB∴⊿AED全等于⊿CFB （角角边定理）∴AE=CF∵AE=CF且AE∥FC∴AECF是平行四边形
