Question

if two equal chords of a circle intersect within the circle prove that the segment of one chord are equal to corresponding segments of the Other cord

Answer 1

Answer:Given; circle with centre O in which chords AB and CD intersect at M.
To Prove: AM = DM and CM = BM



In ΔAOM and ΔDOM
OA = OD (radii)
OM = OM (common side)
So, ΔAOM ≈ Δ DOM
Hence, AM = DM proved
It is given that AB = CD
So, AB – AM = CD – DM
Or, CM = BM proved

Answer 2

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.

solution