**General enunciation:** If two chords of a circle bisect each other, we have to show that their point of intersection is the centre of the circle.

**Particular enunciation:** Suppose two chords AC and BD of the circle ABCD bisect each other at the point O i.e., OA = OC and OD = OB. Let us prove that O is the centre of the circle.

Construction: (A,B) , (B, C), (C, D), and (D, A) are joined.

**Proof:** In ∆ AOD and ∆BOC we get,

OA = OC [given]

OD = OB [given]

And ∠AOD =∠BOD [vertically opposite angle]

∴∆AOD ≅ ∆BOC [by side-angle-side theorem]

Then we get, AD = BC.

Similarly, in ∆ AOB and ∆DOC we can prove that AB = CD.

Since the opposite sides of the quadrilateral ABCD are equal then it is a parallelogram.

∴AB||CD and AD||BC

Now, AB||CD and BD is their secant.

∴∠CBD = ∠ADB ————————–(1) [alternate angles]

Similarly, AD||BC and BD is their secant.

∴∠ABD = ∠CDB ————————–(2) [alternate angles]

Adding equation (1) and (2) we get,

∠CBD+∠ABD = ∠ADB+∠CDB

⟹∠ABC = ∠ADC

∴ ∠B = ∠D

Now ABCD is the quadrilateral inscribed the circle. Then the sum of the opposite angles is two right angles,

i.e., ∠ B+ ∠D = 180^{0}

⟹∠ B+ ∠ B = 180^{0} [∵∠B = ∠D]

⟹2∠ B = 180^{0}

∴ ∠B = 90^{0}

Since one of the angles of the parallelogram ABCD is 90^{0}, hence it is a rectangle and then AB = CD. Hence AC and BD are the two diameters of the circle and they bisect each other at the point O.

∴ O is the centre of the circle. ** (Proved)**