In the figure (ii) given below, chords AB and CD of a circle with centre O intersect at E. If OE bisects ∠AED, prove that AB = CD.

Draw perpendiculars from O to AB and CD.

In △OME and △ONE,

∠OME = ∠ONE = 90°

∠OEM = ∠OEN (Given) = x° (let)

In △OME,

⇒ ∠OME + ∠OEM + ∠MOE = 180°

⇒ 90° + x° + ∠MOE = 180°

⇒ ∠MOE = (90 - x)°

In △ONE,

⇒ ∠ONE + ∠OEN + ∠NOE = 180°

⇒ 90° + x° + ∠NOE = 180°

⇒ ∠NOE = (90 - x)°

∴ ∠MOE = ∠NOE

∴ △OME ≅ △ONE (By A.A.A. axiom)

∴ OM = ON (By C.P.C.T.)

Hence, chords AB and CD are equidistant from the center of circle

Since, chords of a circle that are equidistant from the centre of the circle are equal,

∴ AB = CD.

Hence, proved that AB = CD.