Question

prove that the diagonal of parallelogram bisect each other.....​

Answer 1

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Let consider a parallelogram ABCD in which AB||CD and AD||BC.

In ∆AOB and ∆COD , we have

∠DCO=∠OAB (ALTERNATE ANGLE)

∠CDO= ∠OBA. (ALTERNATE ANGLE)

AB=CD. (OPPOSITE SIDES OF ||gram)

therefore , ∆ AOB ≅ ∆COD. (ASA congruency)

hence , AO=OC and BO= OD. (C.P.C.T)

Answer 2

Given:

• A Parallelogram ABCD in which AB || DC and BC || AD.
• Diagonals AC and BD bisect each other at a point A.

To Prove:

• Diagonals bisect each other i.e OA = OC & OB = OD.

Proof: In ∆AOB and ∆COD , we have

• AB = CD [ Opposite Sides of parallogram are equal ]

• ∠OAB = ∠OCD [ Alternate interior angles as AB || DC and CA cuts them ]

• ∠OBA = ∠ODC [ Alternate interior angles as AB || DC and DB bisects them ]

∴ ∆AOB ≅ ∆COD { By AAS-criteria }

Hence, OA = OC and OB = OD ( by C.P.C.T )