Question

If two equal chords of a circle intersect within the circle , prove that the segments of one chord are equal to the corresponding segments of the other chord.

Answer 1

let PQ and RS be two equal chords of a given circle and they are intersecting each other at point T.
draw perpendiculars OV and OU on these chords.
in triangle OVT and in triangle OUT,
OV=OU (equal chords of a circle are equidistant from the centre)
angle OVT=angle OUT ( each 90')
OT=OT(common)
therefore: triangle OVT is congruent to triangle OUT (RHS congruence rule)
therefore: VT=UT (by CPCT)....(1)
it is given that, 
PQ=RS....(2)
=> 1/2 PQ=1/2 RS
=> OV=RU....(3)
on adding equations (1) and (3), we obtain 
PV+VT=RU+UT
=> PT=RT....(4)
on subtracting equation (4) from equation (2), we obtain
PQ-PT=RS-RT
=> QT = ST.....(5)
equations (4) and (5) indicate that the corresponding segments of chords PQ and RS are congruent to each other.

Answer 2

Given : Let AB and CD be two equal chords of a circle having centre O intersecting each other at point E within the circle.

To Prove :-

(i) AE = CE

(ii) BE = DE

Construction : Draw OM perpendicular at AB, ON perpendicular at CD. Join OE.

Proof :- in rt. angle d ∆s OME and ONE,

angle OME = angle ONE [Each = 90°]

OM = ON

[ because Equal chords are equidistant from the center]

hyp. OE = hyp. OE [Common]

Therefore, By RHS Congruence,

∆OME and ∆ONE are congruent

Therefore, ME = NE ....(1)

Now; O is the centre of circle and OM is perpendicular at AB

Therefore, AM = 1/2 of AB ...(2)

[Because, Perpendicular from the centre bisects the chord]

Similarly, NC = 1/2 of CD ....(3)

But AB = CD [Given]

From (2) and (3),

AM = NC ....(4)

Also, MB = DN .....(5)

Adding (1) and (4),

AM + ME = NC + NE

Hence, AE = CE.

Now, AB = CD (Given)

AE = CE (Proved)

Subtracting AB - AE = CD - CE

Hence, BE = DE.