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