Answer :

Given:- Parallelogram OABC


To Prove:- AC2 + OB2 = OA2 + AB2 + BC2 + CO2


Proof:- Let, O at origin



Therefore,



Distance/length of AC



By triangular law:-


the the vectors form sides of triangle



As AB = OC and BC = OA


From figure



……(i)


Similarly, again from figure





……(ii)


Now,


Adding equation (i) and (ii)


……(iii)


Take RHS


OA2 + AB2 + BC2 + CO2




……(iv)


Thus from equation (iii) and (iv), we get


LHS = RHS


Hence proved


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