Question

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

Answer 1

Next steps
We know that
BE=EM+MB and
DE= DN+ EN
THUS
EM= EN
AND
BE =DE-------(1)
WE ALSO KNOW THAT
AB= AE+ EB
AND
CD =CE+ ED
THEREFORE
BE= DE - - - - - - - - - (FROM(1))
THEREFORE
AE=CE

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.