Question

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

Answer 1

Drop a perpendicular from $O$ to both chords $AB$ and $CD$

In $△OMP$ and $△ONP$

As chords are equal, perpendicular from centre would also be equal.

$OM=ON$

$OP$ is common.

$\angle OMP$$=\angle ONP={90}^o$

$△OMP$ $\cong$ $△ONP$ (RHS Congruence)

$PM=PN$ [By CPCT] ......................(1)

$AM=BM$ (Perpendicular from centre bisects the chord)

Similarly ,$CN=DN$

As $AB=CD$

$AB-AM=CD-DN$

$BM=CN$ .........................(2)

From (1) and (2)

$BM-PM=CN-PN$

$PB=PC$

$AM=DN$ (Half the length of equal chords are equal)

$AM+PM=DN+PN$

$AP=PD$

Therefore , $PB=PC$ and $AP=PD$ is proved.