Theorem – The line drawn through the center of a circle to bisect a chord is perpendicular to the chord

The line drawn through the center of a circle to bisect a chord is perpendicular to the chord.

Given:  A circle with center O, AB is chord of a circle and OC bisect chord at C. i,e. AC = CB.

To prove: OC ⊥ AB

Construction: join OA and OB. 

Proof:  In △ OCA and △ OCB

OA = OB (Radius of circle)

OC = OC (Common side)

AC = CB  (Given)

Hence, △OCA ≅ △OCB (SSS Congruence rule)

∴ ∠OCA = ∠OCB …(1) (Corresponding parts of congruent triangles)

In line AB,  ∠OCA and ∠OCB from linear pair

∴ ∠OCA + ∠OCB = 1800

From (1)

∴ ∠OCA + ∠OCA = 1800

∴ 2∠OCA = 1800

∴ ∠OCA = 180/2

∴ ∠OCA = 900

 Thus, OC ⊥ AB 

                                         Hence proved

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