Results of Equal Chords

Theorem 1: Equal chords are equidistant from the center.2018-11-24_12-30-12

Given: OL = OM

Construction: Join OA and OC

Proof: We know that the perpendicular from the center of a circle to a chord, bisects the chord.

Since OL \perp AB \Rightarrow AL = \frac{1}{2} AB

Similarly, OM \perp CD \Rightarrow CM = \frac{1}{2} CD

Given AB = CD

\Rightarrow \frac{1}{2} AB = \frac{1}{2} CD

\Rightarrow AL = CM

Now consider \triangle OMC and \triangle OLA

AL = CM (proved above)

OA = OC (radius)

\angle OLA = \angle OMC = 90^o

Therefore \triangle OMC \cong \triangle OLA (By RHS criterion)

\Rightarrow OL = OM . Hence proved that equal chords are equidistant from the center.

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Theorem 2 (Converse of Theorem 1): Chords of a circle which are equidistant from the center are equal.2018-11-24_12-30-22

Given: Two chords AB and CD in circle C(O, r)

Also OL = OM where OL \perp AB and OM \perp CD

To Prove: AB = CD

Construction: Join OA and OC

Proof: OL \perp AB (given)

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

\Rightarrow AL = BL

\Rightarrow AL = \frac{1}{2} AB

Similarly OM \perp CD

\Rightarrow CM = DM

\Rightarrow CM = \frac{1}{2} CD

Consider \triangle OAL and \triangle OCM

OA = OC (radius)

\angle OLA = \angle OMC = 90^o

OL = OM (given)

Therefore \triangle OAL \cong \triangle OCM (By S.A.S criterion)

\Rightarrow AL = CM

\Rightarrow \frac{1}{2} AB = \frac{1}{2} CD

\Rightarrow AB = CD

Hence proved that chords of a circle which are equidistant from the center are equal.

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Theorem 3: Equal chords of concurrent circles are equidistant from the corresponding centers.2018-11-24_12-29-59

Given: Two concurrent circles C(O,r) and C(O',r) and AB = CD

Construction: Join OA and O'C

To Prove: OL = O'M where OL \perp AB and O'M \perp CD

Proof: We know that the perpendicular from the center of a circle to a chord, bisects the chord.

AL = BL \Rightarrow AL = \frac{1}{2} AB

Similarly, CM = DM \Rightarrow CM = \frac{1}{2} CD

Now AB = CD (given)

\Rightarrow \frac{1}{2} AB = \frac{1}{2} CD

\Rightarrow AL = CM

Consider \triangle OAL and \triangle O'CM

OA = O'C (congruent circles with radius r )

\angle OLA = \angle O'MC = 90^o

AL = CM (proved above)

Therefore \triangle OAL \cong \triangle OC'M (By RHS criterion)

Therefore OL = O'M i.e. AB and CD are equidistant from O and O' respectively.

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Theorem 4 (Converse of Theorem 3): Chords of congruent circles which are equidistant from their corresponding centers are equal.

2018-11-24_12-29-59

Given: Two circles C(O,r) and C(O',r) . AB and CD are chords such that OL = O'M . OL \perp AB and O'M \perp CD

To Prove: AB = CD

Construction: Join OA and O'C

Proof: Consider \triangle OLA and \triangle O'MC

OL = O'M (given)

OA = O'C (equal radius)

\angle OLA = \angle O'MC = 90^o

Therefore \triangle OLA \cong \triangle O'MC (By RHS criterion)

\Rightarrow AL = CM

\Rightarrow 2 AL = 2 CM

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

\Rightarrow AB = CD

Hence proved that chords of congruent circles which are equidistant from their corresponding centers are equal.

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Theorem 5: Equal chords of a circle subtends equal angles at the center.2018-11-24_12-29-09

Given: AB and CD are two equal chords in circle C(O,r)

To Prove: \angle AOB = \angle COD

Proof: Consider \triangle AOB and \triangle COD

OA = OD (radius)

OB = OC (radius)

AB = CD (given)

Therefore \triangle AOB \cong \triangle COD

\Rightarrow \angle AOB = \angle COD .

Hence proved that equal chords of a circle subtend equal angles at the center.

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Theorem 6: (converse of Theorem 5) If angles subtended by two chords of a circle at the center are equal, then the chords are equal.2018-11-24_12-29-09

Given: AB and CD are chords of a circle C(O,r) such that \angle AOB = \angle COD

To Prove: AB = CD

Proof: Consider \triangle AOB and \triangle COD

OA = OD (radius)

OB = OC (radius)

\angle AOB = \angle COD (given)

Therefore \triangle AOB \cong \triangle COD (By S.A.S criterion)

Therefore AB = CD (corresponding sides off congruent triangles are equal)

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Theorem 7: Equal chords of congruent triangles subtend equal angles at the center.

2018-11-24_12-28-49

Given: Two concurrent circles C(O, r) and C(O', r) . Chords AB = CD

To Prove: \angle AOB = \angle CO'D

Proof: Consider \triangle AOB and \triangle CO'D

OA = O'C (radius)

OB = O'D (radius)

AB = CD (given)

Therefore \triangle AOB \cong \triangle CO'D (By S.S.S criterion)

Hence \angle AOB = \angle CO'D (corresponding angles of congruent triangles are equal).

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Theorem 8 (converse of Theorem 7): If the angles subtended by the two chords of congruent circles at the corresponding centers of circles are equal, the chords are equal.2018-11-24_12-28-49

Given: Two congruent circles C(O,r) and C(O',r) . AB and CD are chords such that \angle AOB = \angle CO'D

To Prove: AB = CD

Proof: Consider \triangle AOB and \triangle CO'D

OA = O'C (radius)

OB = O'D (radius)

\angle AOB = \angle CO'D (given)

Therefore \triangle AOB \cong \triangle CO'D (By S.A.S criterion)

\Rightarrow AB = CD

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Theorem 9: Of any two chords of a circle, the chord which is larger is nearer to the circle.2018-11-24_12-28-19

Given: Two chords AB and CD of a circle with center O such that AB > CD

Construction: Join OA and OC

To Prove: OL < OM or OM > OL

Proof: We know that the perpendicular from the center of a circle to a chord, bisects the chord.

OL \perp AB \Rightarrow AL = \frac{1}{2} AB

And OM \perp CD \Rightarrow CM = \frac{1}{2} CD

In right triangles \triangle OAL and \triangle OCM , we have

OA^2 = OL^2 + AL^2

And OC^2 = OM^2 + CM^2

Since OA = OC (radius)

OL^2 + AL^2 = OM^2 + CM^2

Given AB > CD

\Rightarrow \frac{1}{2} AB > \frac{1}{2} CD

\Rightarrow AL > CM

\Rightarrow AL^2 > CM^2 (Adding OL^2 to both sides)

Therefore OL^2 + AL^2 > OL^2 + CM^2

\Rightarrow OM^2 + CM^2 > OL^2 + CM^2

\Rightarrow OM^2 > OL^2

\Rightarrow OM > OL.

Hence proved that of any two chords of a circle, the chord which is larger is nearer to the circle.

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Theorem 10 (converse of Theorem 9): Of any two chords of a circle, the chord which is nearer is larger.2018-11-24_12-28-19

Given: Two chords AB and CD of a circle C(O, r) such that OL < OM , where OL and OM are perpendicular from O on AB and CD respectively.

To Prove: AB > CD

Proof: We know that the perpendicular from the center of a circle to a chord, bisects the chord.

Therefore AL = \frac{1}{2} AB \ \ \ \& \ \ \ \  CM = \frac{1}{2} CD

In right triangles \triangle OAL and \triangle OCM we have

OA^2 = OL^2 + AL^2

And OC^2 = OM^2 + CM^2

\Rightarrow AL^2 = OA^2 - OL^2 … … … … … i)

And CM^2 = OC^2 - OM^2 … … … … … ii)

Now OL < OM

\Rightarrow OL^2 < OM^2

\Rightarrow = OL^2 > - OM^2 (Adding OA^2 on both sides)

OA^2 - OL^2 > OA^2 - OM^2

AL^2 > OA^2 - OM^2

AL^2 > OC^2 - OM^2

Since OA^2 = OC^2 (radius of the circle)

AL^2 > CM^2

\Rightarrow AL > CM

\Rightarrow 2AL > 2 CM

\Rightarrow AB > CD . Hence proved that of any two chords of a circle, the chord which is nearer is larger.

 

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