Question

prove that the sum of the squares of the diagonals of parallelogram is equal to the sum of the squares of its sides​

Answer 1

answer

let abcd be the parallelogram let diagonals ac and bd interset in o

consider triangle abc

let bc=x and ab=y

by stewart theorem x^2* d/2 + y^2*d/2 =[d/2+d/2][d^2/4+d^2/4]

d/2(x^2+y^2)= d(d^2/2)

x^2+y^2= 2/d*d*d^2/2=d^2

multiply rhs and lhs by 2 to get the result