Prove that the sum of the squares of the diagonals of parallelogram is equal to the sum of the squares of its sides.

Explanation

Given : `() square ABCD` is a parallelogram
`(ii)` Diagonals `AC` and `BD` intersect at `O`.
To prove: `AC^(2)+BD^(2)=AB^(2)+BC^(2)+CD^(2)+AD^(2)`
Proof : `square ABCD` is a parallelogram……..(Given)
`{:( :. CD=AB......(1)),(AD=BC..........(2)):}}` (Opposide sides of parallelogram are equal )
Also `AO=OC=(1)/(2)AC`......`(3)`
`BO=OD=(1)/(2)BD`........`(4)` .........(Diagonals of parallelogram bisect each other)
In `DeltaABC`,
seg `BO` is the median ......(By definition)
`:.` by Apollonius theorem,
`AB^(2)+BC^(2)=2BO^(2)+2OC^(2)`
`:.AB^(2)+BC^(2)=2((1)/(2)BD)^(2)+2((1)/(2)AC)^(2)`........[From `(3)` and `(4)`]
`:.AB^(2)+BC^(2)=2xx(1)/(4)BD^(2)+2xx(1)/(4)AC^(2)`
`:.AB^(2)+BC^(2)=(1)/(2)BD^(2)+(1)/(2)AC^(2)`
Multiplying each term by `2`, we get,
`2AB^(2)+2BC^(2)=BD^(2)+AC^(2)`
`:.AB^(2)+AB^(2)+BC^(2)+BC^(2)=BD^(2)+AC^(2)`
`:.AB^(2)+CD^(2)+BC^(2)+AD^(2)=BD^(2)+AC^(2)`......[From `(1)` and `(2)`]
`:.AB^(2)+BC^(2)+CD^(2)+AD^(2)=BD^(2)+AC^(2)`
i.e. `AC^(2)+BD^(2)=AB^(2)+BC^(2)+CD^(2)+AD^(2)`.