Proof: the square sum of two diagonals of a parallelogram is equal to the square sum of four sides

It is known that in parallelogram ABCD, AC and BD are its two diagonals. This paper proves: ac2 + BD2 = AB2 + BC2 + Cd2 + ad2. It is proved that if AE ⊥ BC is at point E and the extension of DF ⊥ BC is at point F, then ⊥ AEB = ⊥ DFC = 90 °.⊥ quadrilateral ABCD is a parallelogram, ⊥ AB = DC, ab ∥ CD, ⊥ Abe = ⊥ DCF, ≌ Abe ≌ DCF, ≌ AE = DF, be = CF, Ac2 = AE2 + EC2 = AE2 + (bc-be) 2, BD2 = df2 + BF2 = df2 + (BC + CF) 2 = AE2 + (BC + be) 2, ac2 + BD2 = 2ae2 + 2bc2 + 2be2 = 2 (AE2 + be2) + 2bc2, and AE2 + be2 = AB2, that is ac2 + BD2 = 2 (AB2 + BC2). ∵ AB = CD, ad = BC, ∵ ac2 + BD2 = AB2 + BC2 + Cd2 + ad2

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