Circle Theorems on Central Angles and Inscribed Angles

Circle Theorems on Central Angles and Inscribed Angles

Circle Theorems on Central Angles and Inscribed Angles

The angle formed at the centre of a circle is called a central angle and the angle formed at the circumference of a circle is called a circumference angle or inscribed angle.

Central angle and inscribed angle

To understand and prove the circle theorems on central angles and inscribed angles, we use the following axioms on arcs and angles subtended by them:

1.    Degree measurement of a full circle arc is equivalent to 360°.

2.    Central angle of a circle is equal to the degree measurement of its opposite arc.

3.    Inscribed angle of a circle is equal to the half of the degree measurement of its opposite arc.

 

Theorems:

1.    Central angle of a circle is double the inscribed angle standing on same arc.

2.    The angles at the circumference of a circle standing on the same arc are equal.

3.    Inscribed angle on a semi-circle is a right angle.

 

Proofs:

1.   Central angle of a circle is double the inscribed angle standing on same arc.

O is the centre of a circle. ∠AOB is a central angle and ∠ACB is a inscribed angle standing on same arc AB.

Given: O is the centre of a circle. AOB is a central angle and ACB is a inscribed angle standing on same arc AB.

To Prove: AOB = 2ACB

Proof:

               Statements                        Reasons 

1.    AOB = Arc AB --------> Central angle is equal to its opposite arc.

2.    ACB = ½ Arc AB -----> Inscribed angle is equal to the half of its opposite arc.

Or, 2ACB = Arc AB

3.    AOB = 2ACB ------------> From statements 1 and 2.                                                                                                                                                                                                                                                 Proved.                 

 

2.   The angles at the circumference of a circle standing on same arc are equal.

O is the centre of a circle. ∠ABC and ∠ADC are the inscribed angles standing on same arc AC

Given: O is the centre of a circle. ABC and ADC are the inscribed angles standing on same arc AC.

To Prove: ABC = ADC

Proof:

                Statements                        Reasons

1.       ABC = ½ Arc AC ------> Inscribed angle is equal to the half of its opposite arc.

2.       ADC = ½ Arc AC ------> Inscribed angle is equal to the half of its opposite arc.

3.       AOB = 2ACB ---------> From statements 1 and 2.

                                                                                 Proved.                                                                                            

 

3.   Inscribed angle on a semi-circle is a right angle.

O is the centre of a circle. AB is a diameter

Given: O is the centre of a circle. AB is a diameter.

To Prove: ACB = 90°

Proof:

                Statements                        Reasons

1.       AOB = 180° ----------> Being AOB a straight angle.

2.       ACB = ½ AOB ------> Inscribed angle is equal to the half of its opposite arc.

3.       ACB = ½ × 180° --------------> From statements 1 and 2.

4.       ACB = 90° ------------> From statement 3.

                                                                              Proved.                                                                                         

 

Look at the following worked out examples of geometrical problems on circle:

 

Workout Examples

Example 1: Find the value of x and y in the given figure.

Example 1: Circle

Solution: From the figure,

          OCA = OAC = 50° [Being  OA = OC, Radii of same circle]

          y + 50° + 50° = 180° [Sum of  the angels of DOAC]

or,     y + 100° = 180°

or,     y = 180° - 100°

or,     y = 80°

 

          x = ½ of y [Inscribed angle is half of central angle]

             = ½ × 80°

             = 40° 

 

Example 2: Find the value of y in the given figure.

Example 2: Circle

Solution: From the figure,

          ABC = 90° [An angle formed on semi-circle]

          y + 50°  + 90° = 180° [Sum of the angles of DABC]

or,     y + 140° = 180°

or,     y = 180° - 140°

             = 40°         

 

Example 3: Find the value of z in the given figure.

Example 3: Circle

Solution: From the figure,

          Reflex AOC = 2ABC [Central angle is double of inscribed angle on the same arc]

                            = 2 × 130°

                            = 260°

           

          z + reflex AOC = 360° [Angles at one complete rotation]

or,     z + 260° = 360°

or,     z = 360° - 260°

or,     z = 100°

 

Example 4: What is the size of CBX in the given figure.

Example 4: Circle

Solution: Here, ADB = 32° and BXC = 85°

          ACB = ADB = 32° [Inscribed angles on same arc]

           

          CBX + ACB + BXC = 180° [Sum of angles of DBXC]

or,     CBX + 32° + 85° = 180°

or,     CBX + 117°= 180°

or,     CBX = 180° - 117°

or,     CBX = 63°

 

Example 5: In the given diagram, PQR is an isosceles triangle with PQ = PR and a circle is drawn with PQ as a diameter. Prove that QS = RS.

In DPQR, PQ = PR. O is centre and PQ is a diameter of circle

Solution:

Given: In DPQR, PQ = PR. O is centre and PQ is a diameter of circle.

To Prove: QS = RS.

Construction: Points P and S are joined.

Proof:

       

1.    PSQ = 90° --------> Being PSQ is in semi-circle with diameter PQ

2.    In DPSQ and DPSR

i.        PSQ = PSR (R) --------> Each of 90° (Statement 1)

ii.       PQ = PR (H) -------> By given

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

3.    DPSQ DPSR -------------> By RHS axiom

4.    QS = RS -------------> Corresponding sides of triangles.

                                                                                                         Proved.

 

Example 6: In the given figure, O is the centre of the circle, PQ the diameter and AOPQ, prove that PBR = PAO.

PQ is the diameter of a circle with centre at O and AP⊥PQ

Solution:

Given: PQ is the diameter of a circle with centre at O and APPQ.

To Prove: PBR = PAO.

Construction: Points R and Q are joined.

Proof:

      

1.    PRQ = 90° --------> Being PRQ is in semi-circle with diameter PQ

2.    In DPQR and DPAO

i.        PRQ = PAO ------> Being APPQ, and from statement 1.

ii.       QPR = APO -------> Common angle.

iii.     PQR PAO -------> Remaining angles.

3.    PQR = PBR -------------> Being inscribed angles on same arc PR.

4.    PBR = PAO -------------> From statements 1(iii) and 3.

                                                                                                         Proved.

 

Example 7: In the given figure, O is the centre of the circle, AB the diameter and arc BC = arc CD. Prove that AD || OC.

O is the centre of circle. AB is the diameter. Arc BC = arc CD

Solution:

Given: O is the centre of circle. AB is the diameter. Arc BC = arc CD

To Prove: AD || OC.

Proof:

        

1.    BAD = ½ of arc BCD ----> Inscribed angle is equal to the half of its opposite arc

2.    BAD = ½ of arc (BC + CD) -------> whole part axiom from statement 1

3.    BAD = ½ of arc (BC + BC) --------> Being arc BC = arc CD (given).

4.    BAD = ½ × arc 2BC -------------> From statements 3.

5.    BAD = arc BC ------------------> From statement 4.

6.    BOC = arc BC --------------> Central angle is equal to its opposite arc.

7.    BAD = BOC -----------> From statements 5 and 6.

8.    AD // OC ---------> Being corresponding angles equal (statement 7)

                                                                                                         Proved.

 

You can comment your questions or problems regarding circle and theorems on central angle and inscribed angles here.


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