# Perpendicular from the Centre to a Chord

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**Activity :**

Draw a circle on a tracing paper. Let O be its centre. Draw a chord AB. Fold the paper along a line through O so that a portion of the chord falls on the other. Let the crease cut AB at the point M. Then, ∠ OMA = ∠ OMB =

or OM is perpendicular to AB.

Does the point B coincide with A?

Yes, it will. So MA = MB.

Give a proof yourself by joining OA and OB and proving the right triangles OMA and OMB to be congruent. This example is a particular instance of the following result:

**Theorem 9.3**

**Perpendicular Bisector from Centre Theorem:** The perpendicular from the centre of a circle to a chord bisects the chord.

**Given:** OM is a perpendicular from the centre O i.e. ∠OMA = ∠OMB = 90°.

**Construction:** Join OA and OB.

**To Prove:** AM = BM

**Proof:** In the triangles △𝑂𝑀𝐴 and △OMB:

(i) OA = OB as they are

(ii) OM is

(iii) ∠OMA = ∠OMB =

Therefore, by

Thus, AM = BM

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What about the converse of this theorem? Let's prove that as well. The converse is:

**Theorem 9.4**

**Chord Bisector Perpendicular Theorem:** **The line drawn through the centre of a circle to bisect a chord is perpendicular to the chord.**

**Given:** AM = BM

**Construction:** Join OA and OB.

**To Prove:** ∠OMA = ∠OMB = 90°

**Proof:** In the triangles △𝑂𝑀𝐴 and △OMB:

(i) OA = OB as they are

(ii) OM is

(iii) AM = BM (Given)

Therefore, by

Thus, by CPCT: ∠OMA = ∠OMB

We also know that AB is a chord i.e.

2 ∠OMB = 180° ⟹ ∠OMB = 90°