Property: The diagonals of a parallelogram bisect each other .
Given : ABCD is parallelogram , AC & BD intersect at O.
To prove : OA = OC and OB = OD .
Proof : In ∆ AOB and ∆ COD , we know that
∠OAB = ∠ OCD (Alternate angle )
∠OBA = ∠ ODC (Alternate angle )
`bar (AB) = bar (CD)` (opposite sides are equal in parallelogram)
∆ AOB ≅ ∆ COD (ASA Axiom)
Hence OA = OC and OB = OD