Question

prove that the diagonals of a parallelogram bisected each other

Answer 1

Take a parallelogram ABCD.  Let the diagonals intersect at point O.

As  BC || DA, ∠CAD = ∠ACB   (AC is the transversal line).
                      ∠ADB = ∠DBC   (BD is the transversal line).
Also  ∠AOD = ∠BOC.

If we compare the ΔAOD and ΔBOC, we have three corresponding angles. Also BC = DA.  BO & OD are parallel.  Also AO & OC are parallel.  Hence both triangles are congruent.

   Hence  BO = OD    and AO = OC.

So the diagonals BD and AC bisect each other. 

Proved.

Answer 2

Given: A parallelogram ABCD such that its diagonal AC and BD intersect at O.

To prove: OA = OC and OB = OD.

Proof: Since ABCD is a parallelogram. Therefore,

AB || DC and AD || BC

Now, AB || DC and transversal AC intersects them at A and C respectively.

∴ ∠BAC = ∠DCA [∵ Alternate interior angles are equal]

→ ∠BAO = ∠DCO .......(i)

Again, AB||DC and BD intersects them at B and D respectively.

∴ ∠ABD = ∠CDB [∵ Alternate interior angles are equal]

→ ∠ABO = ∠CDO .......(ii)

Now, in △AOB and △COD, we have

∠BAO = ∠DCO [From (i)]

AB = CD [∵ Opposite sides of a parallelogram are equal]

and, ∠ABO = ∠CDO [From (ii)]

So, by ASA congruence criterion,

△AOB ≅ △COD

→OA = OC and OB = OD [c.p.c.t.]

Hence, OA = OC and OB = OD