Question

if two equal chords of a circle intersect within the circle prove that the segment of are chard are equal to corresponding segment of other chord ​

Answer 1

Step-by-step explanation:

Drop a perpendicular from O to both chords AB and CD

In △OMP and △ONP

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

OM=ON

OP is common.

∠OMP=∠ONP=90  o

△OMP ≅ △ONP (RHS Congruence)

PM=PN

AM=BM              (Perpendicular from centere bisects the chord)

Similarly, CN=DN

As AB=CD

AB−AM=CD−DN

BM=CN

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.

Answer 2

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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=90o

➩△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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