# If two equal chords of a circle intersect within the circle, prove that the segments of one chord are equal

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If two equal chords of a circle intersect within the circle, prove that the segments of one chord are equal to corresponding segments of the other chord

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Given : AB and CD are two equal chords of a circle which meet at E.

To prove : AE = CE and BE = DE

Construction : Draw OM ⊥ AB and ON ⊥ CD and join OE.

Proof : In ∆OME and ∆ONE

OM = ON [Equal chords are equidistant]

OE = OE [Common]

∠OME = ∠ONE [Each equal to 90°]

∴ ∆OME ≅ ∆ONE [RHS axiom]

⇒ EM = EN ...(i) [CPCT]

Now AB = CD [Given]

⇒ 1/ 2 AB = 1/ 2 CD

⇒ AM = CN ..(ii) [Perpendicular from centre bisects the chord]

Adding (i) and (ii), we get

EM + AM = EN + CN

⇒ AE = CE ..(iii) Now, AB = CD ..(iv)

⇒ AB – AE = CD – AE [From (iii)]

⇒ BE = CD – CE Proved. 0 votes
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