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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.

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