Chords AB and CD of a circle with centre O intersect at a point E. If OE bisects angle AED, then prove that AB = CD.

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Solution

Draw OM $⊥$ AB and ON $⊥$ CD

In $Δ$ OEM & $Δ$ OEN,

$∠$ OME = $∠$ ONE = $90^{0}$ [OM $⊥$ AB and ON $⊥$ CD]

$∠$ OEM = $∠$ OEN [Given, OE bisect $∠$ AED]

$∠$ MOE = $∠$ NOE [when two angles of two $Δ$ s are equal, the third angle is also equal]

Also, OE = OE [Common]

OM = ON [C.P.C.T]

Chord AB and CD are equidistant from the centre

Therefore, Chord AB = Chord CD

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