Theorem 10.4: The line drawn through the center of a circle to bisect a chord is perpendicular to the chord Class 9

Given : a circle with center O. AB is a chord of circle and OD bisects AB

I.e AD = BD

To prove: ∠ODA = ∠ODB = 90°

Proof : In △OAD and △OBD

⇒ OA = OB ( Both Radius)

⇒ AD = BD (Given)

⇒ OD = OD (Common)

△AOD ⩭ △BOD (by SSS congruency)

⇒∠ADO = ∠BDO ( By CPCT )

In line AB, AOB is a straight line 

⇒ ∠ODA + ∠ODB = 180°

⇒ ∠ODA = ∠ODA = 180°

⇒ 2∠ODA = 180°

⇒ ∠ODA = 180°/2 

⇒ ∠ODA = 90°

∴ ∠ODA = ∠ODB = 90°

⇒Hence proved

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