Theorem – The perpendicular from the center of a circle to a chord bisects the chord

Theorem

The perpendicular from the center of a circle to a chord bisects the chord.

Given: A circle with center O, AB is chord of a circle and OC perpendicular from the center to the chord  AB.

i.e.    OC ⊥ AB 

therefore  ∠OCA and ∠OCB Both angles are 900.  

To prove: AC = CB (C is the mid point of chord AB)                                 

Construction: join OA and OB. 

Proof: In △OCA and △ OCB

∠OCA = ∠OCB (Both angles are 900 given)
OA = OB (Radius of circle)
OC = OC (Common side)

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

∴ AC = CB …(1) (Corresponding parts of congruent triangles)

∴ AC = CB
 Therefore, C is the mid point of chord AB.                                 

Hence proved

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