31) 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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OM=ON
OM=ONOP is common.
OM=ONOP is common.∠OMP=∠ONP=90
OM=ONOP is common.∠OMP=∠ONP=90 o
OM=ONOP is common.∠OMP=∠ONP=90 o
OM=ONOP is common.∠OMP=∠ONP=90 o △OMP ≅ △ONP (RHS Congruence)
OM=ONOP is common.∠OMP=∠ONP=90 o △OMP ≅ △ONP (RHS Congruence)PM=PN ......................(1)
OM=ONOP is common.∠OMP=∠ONP=90 o △OMP ≅ △ONP (RHS Congruence)PM=PN ......................(1)AM=BM
OM=ONOP is common.∠OMP=∠ONP=90 o △OMP ≅ △ONP (RHS Congruence)PM=PN ......................(1)AM=BM Similarly ,CN=DN
OM=ONOP is common.∠OMP=∠ONP=90 o △OMP ≅ △ONP (RHS Congruence)PM=PN ......................(1)AM=BM Similarly ,CN=DNAs AB=CD
OM=ONOP is common.∠OMP=∠ONP=90 o △OMP ≅ △ONP (RHS Congruence)PM=PN ......................(1)AM=BM Similarly ,CN=DNAs AB=CDAB−AM=CD−DN
OM=ONOP is common.∠OMP=∠ONP=90 o △OMP ≅ △ONP (RHS Congruence)PM=PN ......................(1)AM=BM Similarly ,CN=DNAs AB=CDAB−AM=CD−DNBM=CN .........................(2)
OM=ONOP is common.∠OMP=∠ONP=90 o △OMP ≅ △ONP (RHS Congruence)PM=PN ......................(1)AM=BM Similarly ,CN=DNAs AB=CDAB−AM=CD−DNBM=CN .........................(2)From (1) and (2)
OM=ONOP is common.∠OMP=∠ONP=90 o △OMP ≅ △ONP (RHS Congruence)PM=PN ......................(1)AM=BM Similarly ,CN=DNAs AB=CDAB−AM=CD−DNBM=CN .........................(2)From (1) and (2)BM−PM=CN−PN
OM=ONOP is common.∠OMP=∠ONP=90 o △OMP ≅ △ONP (RHS Congruence)PM=PN ......................(1)AM=BM Similarly ,CN=DNAs AB=CDAB−AM=CD−DNBM=CN .........................(2)From (1) and (2)BM−PM=CN−PNPB=PC
OM=ONOP is common.∠OMP=∠ONP=90 o △OMP ≅ △ONP (RHS Congruence)PM=PN ......................(1)AM=BM Similarly ,CN=DNAs AB=CDAB−AM=CD−DNBM=CN .........................(2)From (1) and (2)BM−PM=CN−PNPB=PCAM=DN (Half the length of equal chords are equal)
OM=ONOP is common.∠OMP=∠ONP=90 o △OMP ≅ △ONP (RHS Congruence)PM=PN ......................(1)AM=BM Similarly ,CN=DNAs AB=CDAB−AM=CD−DNBM=CN .........................(2)From (1) and (2)BM−PM=CN−PNPB=PCAM=DN (Half the length of equal chords are equal)AM+PM=DN+PN
OM=ONOP is common.∠OMP=∠ONP=90 o △OMP ≅ △ONP (RHS Congruence)PM=PN ......................(1)AM=BM Similarly ,CN=DNAs AB=CDAB−AM=CD−DNBM=CN .........................(2)From (1) and (2)BM−PM=CN−PNPB=PCAM=DN (Half the length of equal chords are equal)AM+PM=DN+PNAP=PD
OM=ONOP is common.∠OMP=∠ONP=90 o △OMP ≅ △ONP (RHS Congruence)PM=PN ......................(1)AM=BM Similarly ,CN=DNAs AB=CDAB−AM=CD−DNBM=CN .........................(2)From (1) and (2)BM−PM=CN−PNPB=PCAM=DN (Half the length of equal chords are equal)AM+PM=DN+PNAP=PDTherefore , PB=PC and AP=PD is proved.
OM=ONOP is common.∠OMP=∠ONP=90 o △OMP ≅ △ONP (RHS Congruence)PM=PN ......................(1)AM=BM Similarly ,CN=DNAs AB=CDAB−AM=CD−DNBM=CN .........................(2)From (1) and (2)BM−PM=CN−PNPB=PCAM=DN (Half the length of equal chords are equal)AM+PM=DN+PNAP=PDTherefore , PB=PC and AP=PD is proved.solution