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 ΔOVT and ΔOUT,
OV = OU (Equal chords of a circle are equidistant from the centre)
∠OVT = ∠OUT (Each 90°)
OT = OT (Common)
∴ ΔOVT ≅ ΔOUT (RHS congruence rule)
∴ VT = UT (By CPCT) ... (1)
It is given that,
PQ = RS ... (2)
⇒
⇒ PV = 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.