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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.

Q: 2 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: two equal chords of a circle intersect within the circle

To prove: Segments of one chord are equal to corresponding segments of the other chord i.e. $AE = CE$ and $BE=DE$.

Construction: Join OE and draw $O M \perp A B$ and $O N \perp C D$

Proof : 

 

In $\triangle \mathrm{OME}$ and $\triangle \mathrm{ONE}$,
$\mathrm{AE}=\mathrm{AE} \quad$ (Common)
$\mathrm{OM}=\mathrm{ON}$
(Equal chords of a circle are equidistant from the centre)
$\angle \mathrm{OME}=\angle \mathrm{ONE} \quad$ (Both are right-angled)
Thus, $\triangle \mathrm{OME} \cong \triangle \mathrm{ONE}$
(By SAS rule)
$\mathrm{EM}=\mathrm{EN}$  (CPCT)--------1
$A B=C D$ $\qquad$ (Given)-------2
$\Rightarrow \frac{1}{2} A B=\frac{1}{2} C D$

$\Rightarrow A M=C N$------3
Adding 1 and 3, we have
$\mathrm{AM}+\mathrm{EM}=\mathrm{CN}+\mathrm{EN}$
$\Rightarrow A E=C E$
Subtract 4 from 2, we get
$\mathrm{AB}-\mathrm{AE}=\mathrm{CD}-\mathrm{CE}$
$\Rightarrow E B=E D$

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seema garhwal

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