Question

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 .

Answer 1

Given-AB and CD are two equal chords with centre O. AB and CD intersect at P.
To prove-BP=OP
AP=CP
Construction-OM perpendicular AB,OM perpendicular CD and join OP
proof-since OM perpendicular AB
AM=MB=1/2AB (1)
Also,AM perpendicular CD
CN=ND=1/2CD (2)
and AB=CD (3) ( given)
from (1),(2) and (3)
AM=CN and MB=ND(4)
Now triangle OMP and triangle OMP
Angle OMP= angle OMP (90°)
OM = ON ( two equal chords of a circle
are equidistant from the
centre)
OP =OP (common )
By RHS rule
triangle OMP congruent to triangle ONP
PM =PM (CPCT)(5)
Add (4) and (5)
PM +MB =PN =NB
BD=DP
Subtract (4) from (5)
AM -PM= CN-PN
AP =CP

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.