In rectangle ABCD, take any point P, connect AP, BP, CP, DP, ask the relationship of AP, BP, CP, DP

PA^2+PC^2=PB^2+PD^2
Let p be above ad, PF through P, vertical ad to e, vertical BC to F
From Pythagorean theorem PA ^ 2 = PE ^ 2 + AE ^ 2
PC^2=PF^2+CF^2
PB^2=PF^2+BF^2
PD^2=PE^2+DE^2
AE = BF, ed = FC
Then PA ^ 2 + PC ^ 2 = Pb ^ 2 + PD ^ 2

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