Question

If two equal chords of a circle intersect within a circle. Prove that the segment of one chord are equal to the corresponding segment of another.

Answer 1

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

In $△OMP$ and $△ONP$, we have

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

$OM=ON$

$OP$ is common.

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

$△OMP\cong △ONP$ ( by RHS Congruence)

$PM=PN................\left(1\right)$

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

Similarly, $CN=DN$

As $AB=CD$

$AB-AM=CD-DN$

$BM=CN..............\left(2\right)$

From $eq.\left(1\right)$ and $\left(2\right)$, we have

$BM-PM=CN-PN$

$PB=PC$

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

$AM+PM=DN+PN$

$AP=PD$

Therefore, $PB=PC$ and $AP=PD$