Using vector method, prove that if the diagonals of a parallelogram are equal, then it is a rectangle.

Explanation

Let OACB be a parallelogram such that OC =AB
` Let vecOA=veca,vecOB=vecb`
Now OC = AB
` or OC^(2) = AB^(2)`
`or (vecOA+vecAC)^(2)= (vecAO+vecOB)^(2)`
`or (vecOA+vecOB)^(2)=(-vecOA+vecOB)^(2)`
` Rightarrow (veca+vecb)=(-veca+vecb)^2`
`veca^(2)+vecb^(2)+2veca.vecb=veca^(2)+vecb^(2)-2veca.vecb`
`2veca.vecb=-2veca.vecb`
`or 4 veca.vecb=0`
`or veca.vecb=vec0`
Hence, `veca and vecb` are perpendicualar,i.e.,
`angleAOB= 90^(@)`
Therefore, OACB is a rectangle.