Chords and Circle Theorems on Chords

Chords and Circle Theorems on Chords

Chords and Circle Theorems on Chords   

The line segment joining any two points on the circumference of a circle is called the chord of the circle. In the figure given below: AB, CD and EF are the chords. AB is also the diameter. Diameter is the longest chord.

AB, CD and EF are the chords

Circle Theorems on Chords

1.    In a circle, a perpendicular from the centre to its chord, bisect the chord.

2.    Equal chords of a circle are equidistant from the centre of the circle.

Proofs:

1.   In a circle, a perpendicular from the centre to its chord, bisect the chord.

Given: O is the centre of circle and OM is perpendicular to chord AB.

           Given: O is the centre of circle and OM is perpendicular to chord AB.

           To Prove: AM = BM

           Construction: OA and OB are joined. 

           Proof:

                    Statements                            Reasons

 1.    In DAOM and DBOM

 i.     AMO = BMO (R) --------> Both are right angles; OMAB

 ii.    OA = OB (H) --------------> Radii of the same circle

 iii.   OM = OM (S) --------------> Common side

 2.    DAOM DBOM -----------------> By RHS axiom

 3.    AM = BM -----------------> Corresponding sides of congruent triangles.

                                                                                               Proved.

Note: The converse of this theorem i.e. “A straight line joining the centre of circle and mid-point of a chord is perpendicular to the chord.” is also true.

 

2.   Equal chords of a circle are equidistant from the centre of the circle. 

           Given: O is the centre of the circle, chord AB = chord CD. OMAB and ONCD.

           To Prove: OM = ON

           Construction: OA and OC are joined. 

           Proof:

                     Statements                           Reasons

 1.    AM = AB/2 and CN = CD/2 ---> Perpendicular from centre to chord bisect it.

 2.    AB = CD ---------------------> By given

 3.    AM = CN --------------------> From statements 1 and 2

 4.    In DAMO and DCNO

  i.       AMO = CNO (R) ---------> Both are right angles

  ii.     OA = OC (H) ---------------> Radii of the same circle

  iii.   AM = CN (S) ---------------> From statement 3.

 5.    DAMO DCNO ------------------> By RHS axiom

 6.    OM = ON -----------------> Corresponding sides of congruent triangles.

                                                                                              Proved.

Note: The converse of this theorem i.e. “Chords which are at equidistant from the centre of a circle are equal in length.” is also true.


Look at the following worked out examples:

 

Workout Examples

Example 1: In the figure, O is the centre of the circle and OMAB. If OP = 10cm and OM = 6cm, find the length of the chord AB.

Example 1: In the figure, O is the centre of the circle and OM⊥AB. If OP = 10cm and OM = 6cm, find the length of the chord AB.

Solution: In the figure,

          OP = 10cm

          OA = 10cm [ Radii of same circle]

          OM = 6cm

In right angle triangle AOM, AM = √(〖OA〗^2-〖OM〗^2 ) = √(〖10〗^2-6^2 ) = √(100-36) = √64 = 8cm

          Now,

                   AB = 2 × AM [ Perpendicular bisects the chord]

                         = 2 × 8cm

                         = 16cm

 

Example 2: In the figure, O is the centre of the circle, chord AB = 6cm, chord CD = 12cm, OMCD and ONAB. Find the distance between the chords if the radius is 3√5cm.

Example 2: In the figure, O is the centre of the circle, chord AB = 6cm, chord CD = 12cm, OM⊥CD and ON⊥AB. Find the distance between the chords if the radius is 3√5cm.

Solution: In the figure,

          OP = OC = 3√5cm

          CM = CD/2 = 12/2 = 6cm

          AN = AB/2 = 6/2 = 3cm         

In right angle triangle AON, ON = √(〖OA〗^2-〖AN〗^2 ) = √(〖(3√5)〗^2-3^2 ) = √(45-9) = √36 = 6cm  and  In right angle triangle COM, OM = √(〖OC〗^2-〖CM〗^2 ) = √(〖(3√5)〗^2-6^2 ) = √(45-36) = √9 = 3cm

          Now,

                   MN = ON - OM

                         = 6 – 3 cm

                         = 3cm

          The distance between chords is 3cm.

 

Example 3: In the fugure O is the centre of the circle and DABC DDBC. Prove that OM is the bisector of AMD.

Example 3: In the fugure O is the centre of the circle and DABC ≅ DDBC. Prove that OM is the bisector of ∠AMD.

Solution:

Given: In the figure O is the centre of the circle and DABC DDBC.

To Prove: OM is the bisector of AMD.

Construction: OPAC and OQBD are drawn.

Proof:

       Statements                                 Reasons

1.    AC = BD ---------> Corresponding sides of Ds ABC and BCD

2.    OP = OQ --------> Equal chords of a circle are equidistant from the centre.

3.    In DOPM and DOQM

i.        OPM = OQM (R) ---------> Both are right angles

ii.      OM = OM (H) ----------> Common side

iii.     OP = OQ (S) -----------> From statement 2

4.    DOPM DOQM -------------> By RHS axiom

5.    PMO = QMO i.e.

AMO = DMO ------------> Corresponding angles of triangles.

6.    OM is the bisector of AMD --------------> From statement 5.

                                                                                  Proved.

 

Example 4: In the fugure O is the centre of the circle and PM = QN. Prove that DPOQ is an isosceles triangle.

Example 4: In the fugure O is the centre of the circle and PM = QN. Prove that DPOQ is an isosceles triangle.

Solution:

Given: O is the centre of the circle and PM = QN.

To Prove: DPOQ is an isosceles triangle.

Construction: OM and ON are joined.

Proof:

       Statements                                 Reasons

1.   OMN = ONM --------> Base angles of an isosceles DOMN (OM = ON)

2.   OMP = ONQ --------> Supplementary angles of equal angles OMN =                                                      ONM (Statement 1)

3.   In DOMP and DONQ

i.        OM = ON (S) ---------> Radii of same circle

ii.       OMP = ONQ (A) ----------> From statement 2

iii.     PM = QN (S) -----------> By given

4.   DOMP DONQ -------------> By SAS axiom

5.   OP = OQ -------------> Corresponding sides of triangles.

6.   DPOQ is an isosceles triangle --------> From statement 5.

                                                                                           Proved.

 

Example 5: In the fugure, chords AB and CD intersect at M. If OM is the bisector of BMD, prove that AB = CD.

Example 5: In the fugure, chords AB and CD intersect at M. If OM is the bisector of ∠BMD, prove that AB = CD.

Solution:

Given: OM is the bisector of BMD.

To Prove: AB = CD.

Construction: OXAB and OYCD are drawn.

Proof:

       Statements                                 Reasons

1.    In DOMX and DOMY

i.        OM = OM (S) ---------> Common side

ii.       OMX = OMY (A) -------> Being OM a bisector of BMD (given)

iii.     MXO = MYO (A) -------> Both right angles.

2.    DOMX DOMY -------------> By SAA axiom

3.    OX = OY -------------> Corresponding sides of triangles.

4.    AB = CD ------> Chords equidistant from centre of a circle are equal.

                                                                                                     Proved.

 

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