# 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.

S seema garhwal

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. AP = CP and BP=DP.

Construction : Join OP and draw $OM\perp AB\, \, \, \, and\, \, \, ON\perp CD.$

Proof :

In $\triangle$OMP and $\triangle$ONP,

AP = AP         (Common)

OM = ON          (Equal chords of a circle are equidistant from the centre)

$\angle$OMP = $\angle$ONP      (Both are right angled)

Thus,  $\triangle$OMP $\cong$ $\triangle$ONP         (By SAS rule)

PM = PN..........................1 (CPCT)

AB = CD ............................2(Given )

$\Rightarrow \frac{1}{2}AB=\frac{1}{2}CD$

$\Rightarrow AM = CN$......................3

Adding 1 and 3, we have

AM + PM = CN + PN

$\Rightarrow AP = CP$

Subtract 4 from 2, we get

AB-AP = CD - CP

$\Rightarrow PB = PD$

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