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If two equal chords of a circle inter...

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 Consider AB and CD are two equal chords of a circle, which meet at point E.
To prove AE=CE and BE=DE
Construction Draw `OM bot AB and ON bot CD` and join OE where O is the centre of circle.

Proo In `Delta OME and DeltaONE`,
OM=ON [equal chords are equidistant from the centre]
OE=OE [common side]
and `angleOME=angleONE ["each" 90^(@)]`
`:. DeltaOME cong DeltaONE` [by RHS congurence rule]
`rArr EM=EN` [by CPCT]...(i)
Now, AB =CD
On dividing both sides by 2, we get
` (AB)/2=(CD)/2 rArr AM=CN` ...(ii)
[ since, perpendicular drawn centre of circle to chord bisects the chord i.e., AM =MB and CN=ND]
On adding Eqs. (i) and (ii), we get
EM+AM=EN+CN
`rArr AE+CE` ....(iii)
Now, AB=CD
On subtracting both sides by AE, we get
AB-AE=CD-AE
`rArr BE=CD-CE` [form Eq. (iii)]
`rArr BE=DE`
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