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Home / Geometry / Prove that if two chords of a circle bisect each other, their point of intersection is the centre of the circle.

# Prove that if two chords of a circle bisect each other, their point of intersection is the centre of the circle.

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.

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,

∴ ∠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)

## If two triangles have the three sides of the one equal to the three sides of the other, each to each, then they are equal in all respects.

If two triangles have the three sides of the one equal to the three sides ...