It is proved that the sum of squares of the diagonals of a parallelogram is equal to the sum of squares of its sides

It is proved that the sum of squares of the diagonals of a parallelogram is equal to the sum of squares of its sides

Method 1: using cosine theorem
In the parallelogram ABCD, there are: ab = DC, ad = BC, ∠ a = 180 °～ B, ｜ cosa = - CoSb
According to the cosine theorem, there are:
AC^2＝AB^2＋BC^2－2AB×BC×cosB,······①
BD^2＝AD^2＋AB^2－2AD×AB×cosA＝AD^2＋DC^2＋2BC×AB×cosA.······②
① + 2: AC ^ 2 + BD ^ 2 = AB ^ 2 + BC ^ 2 + ad ^ 2 + DC ^ 2
Method 2: using vector dot product
In the parallelogram ABCD, there are:
Vector AC = vector AB + vector ad, vector BD = vector Ba + vector BC = - vector AB + vector ad
{AC ^ 2} = AB ^ 2 ＋ ad ^ 2 ＋ 2 vector ab · vector ad, ·········································································
BD ^ 2 = ad ^ 2 + AB ^ 2-2 vector ad · vector ad ············································································ ④
③ + 4: AC ^ 2 + BD ^ 2 = AB ^ 2 + ad ^ 2 + ad ^ 2 + AB ^ 2
Obviously, there are ad = BC, ab = DC, AC ^ 2 + BD ^ 2 = AB ^ 2 + BC ^ 2 + ad ^ 2 + DC ^ 2
Method 3: using Pythagorean theorem
Without losing generality, assume that in the parallelogram ABCD, a is an acute angle
Through a and B to DC, the vertical lines are e and f respectively
It is easy to get the following results: AE = BF, ed = FC, EC = ed + DC = FC + DC, DF = dc-fc
According to Pythagorean theorem, there are: AC ^ 2 = AE ^ 2 + EC ^ 2, BD ^ 2 = BF ^ 2 + DF ^ 2
AC^2＋BD^2＝2BF^2＋（FC＋DC）^2＋（DC－FC）^2＝2BF^2＋2FC^2＋2DC^2.
According to the Pythagorean theorem, BF ^ 2 + FC ^ 2 = BC ^ 2, AC ^ 2 + BD ^ 2 = 2BC ^ 2 + 2dc ^ 2
There are ad = BC, ab = DC,
∴AC^2＋BD^2＝AB^2＋BC^2＋AD^2＋DC^2.