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 = 1800
⟹∠ B+ ∠ B = 1800 [∵∠B = ∠D]
⟹2∠ B = 1800
∴ ∠B = 900
Since one of the angles of the parallelogram ABCD is 900, 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)