Question 420398
Suppose we have a rhombus, with diagonals AC and BD intersecting at E.


{{{drawing(200, 200, 0, 10, 0, 10,


line(2,2,7,2),
line(2,2,5,6),
line(7,2,10,6),
line(5,6,10,6),
line(2,2,10,6),
line(7,2,5,6),
locate(2,2,A),
locate(5,6,B),
locate(10,6,C),
locate(7,2,D),
locate(6,4,E)

)
}}}


Let AE = EC = x, BE = ED = y, and AB = z. The diagonals of a rhombus are perpendicular, so {{{x^2 + y^2 = z^2}}}. The sum of the squares of the diagonals is given by {{{(2x)^2 + (2y)^2}}}. This is equal to {{{4x^2 + 4y^2 = 4(x^2 + y^2) = 4z^2}}}, which is also equal to the sum of the squares of the side lengths (since each side has length {{{z}}}).