Question

Prove that in a parallelogram with all four sides equal, the diagonals are perpendicular bisector of each other.

Answer 1

Given: A parallelogram ABCD with AB = BC = CD = DA

Also, diagonals AC and BD intersect each other at point O.

To prove: AO and BD are perpendicular bisectors of each other.

Proof:

In ΔAOD and ΔCOB:

∠ADO = ∠OBC (Alternate interior angles as AD || BC)

AD = BC

∠DAO = ∠BCO (Alternate interior angles as AD || BC)

As one side and two angles of a triangle are equal to the one side and two angles of the other triangle, ΔAOD ≅ ΔCOB

Corresponding parts of congruent triangles are also congruent.

⇒ AO = CO and DO = BO

Now, in ΔAOD and ΔAOB:

AO = AO (Common side)

DO = BO (Proved above)

AD = AB (Given)

As all the sides of one triangle are equal to all the sides of the other triangle, ΔAOD ≅ ΔAOB

Corresponding parts of congruent triangles are also congruent.

⇒ ∠AOD = ∠AOB

We know that sum of angles forming a linear pair is 180°.

∴ ∠AOD + ∠AOB = 180°

⇒ 2∠AOD = 180°

⇒ ∠AOD = 90°

Similarly, ∠AOB = ∠BOC = ∠COD = 90°

Thus, diagonals of a parallelogram with all sides equal are perpendicular bisectors of each other.