To show that diagonals of a parallelogram bisect each other.
how to prove it
Given: ABCD is a parallelogram in which AD || BC and AB || CD.
To prove: OA = OC and OB = OD
In ΔAOD and ΔBOC, we have
∠B = ∠D [Alternate interior angles]
AD = BC [Opposite sides of a parallelogram are equal]
and ∠C = ∠A [Alternate interior angles]
⇒ ΔAOD ≅ Δ BOC [ASA congruency]
⇒ OA = OC and OB = OD [c.p.c.t]
PROVE:OA=OC & OB=OD
ABCD is a parallelogram ,therefore
AB//DC and AD//BC
in triangle AOB and COD
<BAO=<DCO( alt interior angles)
AB=CD(opp sides of parallelogram are equal)
<ABO=<CDO(alternate interior angles)
therefore triangle AOB= triangleCOD(ASA congurence criterian)
OA=OC & OB=OD(cpct)
hence, prove that diangonals of parallelogram bisect each other
Theorem 1: If the diagonals of a quadrilateral bisect each other then the quadrilateral is a parallelogram.
Proof: Let the quadrilateral be ABCD with diagonals AC and BD intersecting at P:
Then AP = PC and DP = PB since the diagonals bisect each other. Now consider triangles APD and CPB. The vertical angles APD and CPB are equal, and we have pairs of sides that are equal (AP = PC and DP = PB), so these triangles are congruent by SAS. As a consequence, the corresponding angles, DAP and BCP are equal. But these are alternate interior angles for lines AD and BC with transversal AC. So this proves sideAD is parallel to side BC.
Similarly, if we consider triangles APB and CPD, they are congruent by the same reasoning, so corresponding angles BAP and DCP are equal. These are alternate interior angles for lines AB and DC, so those lines are also parallel.
Therefore ABCD is a parallelogram, by definition.