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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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.
∠OMP=∠ONP=90
o
△OMP ≅ △ONP (RHS Congruence)
PM=PN ......................(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.
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