**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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