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Home / Geometry / CIRCLE | Part-3

CIRCLE | Part-3

Exercise: 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.

General enunciation: if two equal chords of a circle intersect within the circle, it is required to prove that the segments of the one chord.

Screenshot_4

Particular enunciation: Let in the circle ABCD with centre O, AB and CD are two equal chords. They intersect each other at E. it is required to prove that, AE=CE and BE=DE.

Construction: We draw the perpendicular OP and OQ from the centre O to the chords AB and CD respectively.

Proof: O be the centre and OP⊥AB

∴AP=BP

We know that the perpendicular from the centre of a circle to a chord bisects the chord.

i.e. BP= ½ AB

again O be the centre and OQ⊥CD

∴CQ=DQ

i.e. DQ= ½ CD

now AB=CD   [supposition]

or, ½ AB =  ½ CD

∴BP=DQ  ———————— (1)

The distances of the chords AB and CD from the centre O are OP and OQ respectively and AB=CD

∴OP=OQ  [equal chords of a circle are equidistant from the centre]

Now in the right angled ∆OPE and OQE

Hypotenuse OE= hypotenuse OE    [common side]

And OP=OQ

So ∆OPE≌∆OQE

∴ PE=QE ————————– (2)

BP+PE=DQ+QE  [from 1 and]

Or, BE= DE ————————(3)

AB= CD

Or, AB-BE = CD-BE       [from 3]

Or,AB-BE=CD-DE

Or,AE=CE ————————–(4)

Therefore AE=CE and BE=DE  [proved]   [from 3 and 4]   (Proved)

 Exercise: prove that the bisecting points of equal chords lie on a circle

General enunciation: To prove that the bisecting points of equal chords lie on a circle.

Screenshot_5

Particular enunciation: let O be the centre of the circle ABCD. AB, CD and EF are three chords and equal to each other. M, P and N are the bisecting points of AB, CD and EF respectively.

We have to prove that M, N and P lie on a circle.

Construction: We join O, M; O,P and O,N.

Proof: O be the centre of the circle and M, P and N are the midpoints of AB,CD and EF respectively.

The line segment joining the centre of a circle to midpoint of a chord other than diameter is perpendicular to the chord.

∴OM⊥AB, OP⊥CD and ON⊥EF.

The distance of AB, CD and EF from the centre O be OM, OP and ON respectively.

Again, AB= CD=EF

∴OM=OP=ON

If with centre at O and radius equal to OM or ON or OP a circle is drawn, the circle passes through the points M, P and N. Therefore, M,N and P lie on the circle.   [proved]

Ex: Shaw that equal chords drawn from the end points on opposite sides of a diameter are parallel.

General enunciation: We have to show that equal chords drawn from the end points on opposite sides of a diameter are parallel.

Particular enunciation: Let in the circle of centre O, AD is a diameter and AB and CD are two equal chords opposite the diameter.

We have to show that AB∥CD

Screenshot_7

Construction: We drawn the perpendicular OM and ON from the centre O to AB and CD respectively.

Proof: In ∆OAM and ∆ODN

OM=ON  [equal chords of a circle are equidistant from the centre]

OA=OD  [radius of same circle]

And AM=DN  [the midpoints of equal chords are M and N]

So ∆OAM≌∆ODN  [SSS theorem]

∴∠OAM=∠ODN

The angles ∠OAM and ∠ODN produced by the transversal AD of AB and CD are equal and these are alternate angles.

Therefore AB∥CD . (proved)

Exercise: Show that parallel chords drawn from the end points of a diameter are equal.

General enunciation: To show that parallel chords drawn from the end points of a diameter are equal.

Screenshot_8

Particular enunciation: Let in the circle with centre O, AD is a diameter . AB and CD are two parallel chords opposite the diameter AD.

We have to prove that AB=CD.

Construction: We join O,B and O,C.

Proof: In ∆OAB and ∆OCD

AO=OD  [radius of same circle]

BO=OC    [radius of same circle]

And included ∠AOB= included ∠COD   [vertically opposite angle]

So ∆AOB≌∆OCD  [SAS theorem]

∴AB=CD [proved]

 

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